First Semester, Academic Year 2020-2021

 
MODULES for PHYSICS 1-Mechanics and Waves

And for

PHYSICS 1- Physics for Engineering Technology

(Module 1-4)

Prepared By: Prudence Peace V. Bustamante, MS

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 MODULE 1

Units and Physical Quantities

Time frame: 4hrs

Objective: By studying this chapter you will learn:

 Determine the three fundamental quantities of physics and the units physicists use to
measure them.
 Determine how to keep track of significant figures in your calculations.

Introduction:

Physics is one branch of science which deals with matter and energy, with the principles
that govern the motion of particles and waves, with the interaction of particles, and with the
properties of molecules, atoms, atomic nuclei, and of large scale systems such as gases, liquids
and solids.

Pre-test: This pre-test will be conducted through online or face to face oral recitation.

Learning Activities

A. Content/Discussion

Standards and Units

Physics is an experimental science in which experiments are performed to

measure physical quantities. Physical quantities can be expressed in terms of a small number
of fundamental quantities, such as mass, distance, times, electric charges and temperature.
Table 1.1 shows the corresponding SI (International System) units of these quantities. There are
also units not listed in the International System describing the same physical quantity. Take for

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example the unit of length; it can be expressed in terms of meter, inch, yard, mile, etc. In the
conversion between different units of measurement for the same quantity, conversion factors
are used. Table 1.2 shows the commonly used conversion factors. Measurement of very large
or very small quantities makes use of prefixes corresponding to different orders of magnitude.
Table 1.3 shows commonly used prefixes with their corresponding magnitude.

Table 1.1: Basic units of the SI system.

Quantity Unit Name Symbol

Length /Distance Meter M
Mass Kilogram Kg
Time Second s( sec)
Electric Charge Couloumb C
Temperature Kelvin K

Table 1.2: Conversion Factors

1 in 2.54 cm
1 km 0.621 mi
1 liter 1,000 cm3
1 atomic mass unit 1.660 x 10- 27 kg
1N 0.2248 lb
1J 0.239 cal
1 atm 1.013 x 105 Pa

Table 1.3: Unit Prefixes

Prefix Abbreviation Value

Giga G 109
Mega M 106
Kilo K 103
Hecto H 102
Deka D 101
Deci d 10-1

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 Centi c 10-2
Mili m 10-3
Micro µ 10-6
Nano n 10-9

UNCERATAINTY AND SIGNIFICANT FIGURES

In measuring physical quantities experimentally, uncertainties are always present. This

would depend on the technique of measurement and measuring device. Uncertainty is also
called “error”, because in indicates the difference from the measured value and the true value.

On the other hand, accuracy of the measurement indicates how close the measured
value is to the true value. It is usually indicated by a stated uncertainty [by the symbol ± or by
using ( ). For example, the length of a stick is 58.96 ± 0.02 m means that the true value is
between 58.94 m and 58.98 m while the number 1.6454 (21) means. 1.6454 ± 0.0021. The
numbers in the parentheses show the uncertainty in the final digits of the main number. Other
measurement express accuracy in terms of fraction error or percent error, where for example
100 ± (0.05)(100) = 100 ± 5.

What is the equivalent of 90 kilometres per hour in meters per second? In miles per
hour?

Solution:
To convert 90 km/hr in to m/s, we need the following conversion factors: 1000 m / 1
km; 1 hr / 60 min.
and 1 min/ 60 s.

1hr
= (
90 km 90 km
1 hr )( 1000
1 km
m
)( 60 min 60 s )
1 hr
)( 1 min
=25 m/ s

To convert 90 km/hr. we need the conversion factor 1 mi / 1.61 km.

1hr
= (
90 km 90 km
1 hr )( 1.611 mikm )=55.9 mi /hr
Sometimes uncertainty is not stated explicitly. Instead, the uncertainty is indicated by
the number of significant figures. For example we measured the length of this book to be 29.7

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cm, which has three significant figures. Automatically the last digit of 29.7 cm is uncertain, so
the uncertainty is about 0.1 cm.

Rules in Determining the Number of Significant Figures:

1. All nonzero digits are significant. For example, 1894 has four significant figures and 3.67
has three significant figures
2. Zeros appearing between nonzero digits are significant. For example, 30.9 has three
significant figures and 78009 has five significant figures.
3. Zeros appearing before nonzero digits not significant. For example 0.00876 has three
significant and 0.0008 has one significant figure.
4. Zeros at the end of a number without a decimal point may or may not be significant. 3700
could have between two of four significant figures.

What is the number of significant figures of 300.00?

There are five significant figures. Remember that zeros at the end of a number and to the
right of a decimal point are significant.

To resolve the ambiguity in rule number 5, we can express 3700 as 3.7 x 103 to indicate
that it has two significant figures, or we can express 3700 as 3.700 x 103 to show that it has four
significant figures. We can show significant figures much more easily by using scientific
notation, which expresses very large or very small numbers as a number between 1 and 10
multiplied by the appropriate power of 10. Consider the number 0.0000084 m/s. We move the
decimal point 6 places to the right, because we want 0.0000084 m/s to become 8.4 x 10-6 m/s.
Now it is clear that it has two significant figures.

Using the scientific notation, compute 120 x 6000.

Solution:
(1.20 x 102)(6.00 x 103)=(1.20)(6.00) x 102+3 = 7.20 x 105

When multiply or dividing numbers, the number of significant figures in the result can be
no greater than the factor with the lowest number of significant figures. While in adding and

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subtracting numbers, the location of the decimal point matters than the [number of significant
figures.

To understand it better, consider these three numbers: 1.1 , 2.20 , 3.10. Take note that
1.1 , which has two significant figures, has the least number significant figures among the three
numbers. Then the product of the three numbers: 1.1 x 2.20 x 3.10 = 7.5 should also have two
significant figures. However when adding these three numbers, we do not care about the
number of significant figures. Then the sum of the three numbers: 1.1 + 2.20 + 3.10 = 6.4
should also have one digit right after the decimal point since the addend of least number of
digits after the decimal point is 1.1 .

Find the sum of 1.00 and 0.2134.

Solution:
The First number 1.040 has only three significant figures beyond the decimal point,
whereas the second, 0.2134 has four. So the sum can only have three significant figures
beyond the decimal point. Hence,
1.040 + 0.2134 = 1.253

A “liter” is the volume of a cube that is 10 cm x 10 cm x 10 cm in dimensions. Find the

volume of a liter in cubic centimetres and in cubic meters.

Solution:
The volume V of a cube of side L is L3. Therefore,
V = L3 = ( 10 cm)(10 cm)(10 cm) = 103 cm3
To convert 103 cm3 to cubic meters, we need the conversion factor 10-2 m / 1 cm

103 cm3 = (103 cm3) ( )( )( )

10−2 10−2
1 cm 1 cm
10−2
1 cm
=(103)(10-6)m3 = 10-3m3

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B. Performance Task:
I. Problem Solving: Show your solutions completely and neatly.
1. The tallest tree in the world is Sequoia in California that is 368 ft. How high is this in
meters? In Kilometres?
2. A “board foot” is a unit of lumber measure that corresponds to the volume of a piece of
wood 1 ft. square and 1 inch thick. How many cubic inches are there in a board foot?
How many cubic feet? How many cubic centimetres?
3. Water emerges from the nozzle of a fountain in Arizona a 75 km/h and reaches a height
of 170 m. express the water speed in meters per second and in miles per hour and the
height in feet?
4. What is the percent error in each of the following approximation of π ? (a) 22/7 (b)
355/113
5. The mass of the earth is 5.98 x 10 24
kg, and its radius is 6.38 x 108 m. Compute the
density of the earth using powers-of-ten notation and the correct number of significant
figures.

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Exercise – MODULE 1
Name:__________________________________________________________Section:_____

II. Modified Multiple Choices: Write the letter of your choices on the spaces provided.
Then explain or show the solution that justifieth your answer. Each item is worth 2
points; correct answers but without solution will gain only 1 point.
_____1. The measurement 43.1040 has ______ significant figures.
a) Three b) six c) five d) four

_____2. The measurement 5.130 x 10 -4 has _____ significant figures

a) four b) one c) three d) seven

4
(3 x 10¿¿ 8)(8 x 10 )
_____3. Compute ¿
(6 x 10¿ ¿5)¿
a) 6x 107 b) 8x 1017 c) 12x 10-5 d) 1.2 x 10-1

_____4. Evaluate (4.0 x 10-6)(3.0 x 104)

a) 12 x 1010 b) 1.2 x 10-10 c) 12 x 10-5 d) 1.2 x 10-1

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_____5. Which of the following prefixes does NOT represent a fractional part of a whole unit?
a) kilo b) micro c) nano d) deci

_____6. The density of seawater was measured to be 1.07 g/cm3. This density in SI unit is
a) 1.07 x 103 kg/m3 b) 1.07 x 103 kg c) 1.07 x 10-3 kg d) 1.07
x 10-3 kg/m3

_____7. The prefix “mega” means

a) 10-4 b) 106 c) 10-3 d) 10-6

_____8. (2.78 x 10-8) –(5.31 x 10-9)=

a) 2.25 x 10-8 b) 22.5 x 10-6 c)2.25 z 10-6 d) 2.52 x 10-8

_____9. Which of the following is a fundamental unit of the SI system of units?

a) kilometer b) kilogram c) gram d) newton

_____10. Light travels a 3 x 108 m/s, and the size of a proton is about 1 fm. Calculate the order
of magnitude
for the time taken for light to pass across a proton.
a) 10-23 s b) 10-7 s c) 10-8 s d) 10-22 s

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Summary

Physical quantities and units: Three

fundamental physical quantities are mass,
length, and time. The corresponding basic SI
units are the kilogram, the meter, and the
second. Derived units for other physical
quantities are products or quotients of the
basic units. Equations must be dimensionally
consistent; two terms can be added only
when they have the same units.
Significant figures: The accuracy of a
Significant figures in magenta
measurement can be indicated by the
number of significant figures or by a stated C 0.424 m
π= = =3.14
uncertainty. The result of a calculation usually 2 r 2( 0.06750 m)
123.62+8.9=132.5
has no more significant figures than the input
data. When only crude estimates are
available for input data, we can often make
useful order-of-magnitude estimates.

POST TEST: This post-test can be conducted through online or face to face.

References
1. Kramer, Laird.Young, Hugh D. (2012) Study guide, Sears & Zemansky's University
physics, 13th edition, Young and Freedman /San Francisco, CA : Pearson.

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 MODULE 2

Vectors and straight-line motion

Time frame: 5hrs

Objective: By studying this chapter you will learn:

 Determine the difference between scalars and vectors, and how to add and subtract
vectors graphically.
 Determine what the components of a vector are, and how to use them in calculations.
 Determine what unit vectors are, and how to use them with components to describe
vectors.
 Describe straight-line motion in terms of average velocity, instantaneous velocity,
average acceleration, and instantaneous acceleration.
 Solve problems involving straight-line motion with constant acceleration, including free-
fall problems.

Introduction:
Some physical quantities can be described by a single number with a unit. However,
many other physical quantities in physics have a direction associated with them. It plays an
important role in many topics of physics, including motion. Motion well be discussed in the latter
part of this module. We will consider a straight-line motion with constant acceleration.

Pre-test: This pre-test will be conducted through online or face to face oral recitation.

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Learning Activities

A. Content/Discussion

VECTOR AND VECTOR ADDITION

When a physical quantity is described by a single number with a unit, we call it a scalar
quantity such as time t , temperature T , mass m , and density ρ , etc. In contrast, a vector
quantity has both a magnitude (the “how much” or “how big” part) and a direction in space
like displacement, velocity ⃗v acceleration a⃗ , etc.

We usually represent a vector quantity such as displacement d⃗ by a single letter with

an arrow above them, such as in Fig. 2.1. In some books they print vector symbols in boldface
italic type with an arrow above them (⃗
A ). A vector can also be represented graphically,
by drawing a line with an arrow on its tip Figure 2.1, the length of the line shows the
vector’s magnitude, and the direction of the line shows the vector’s direction, while its
scale length shows its magnitude. The “magnitude of vector d⃗ ” is simply represented
symbolically by |d⃗| or simply d like other scalar quantities.

If two ⃗
A and ⃗
B vectors have the same direction, they are parallel even if they do not
have the same magnitude. When two vectors ⃗
A and ⃗
B have opposite directions, whether their
magnitudes are the same or not, we say that they are antiparallel. If they have the same
magnitude and the same direction, they are equal, no matter where they are located in space.

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 Figure 2.1 Displacement as a vector quantity. A displacement is always a straight-line
segment directed from the starting point to the ending point, even if the path is curved

[1].

We define the negative of a vector as a vector having the same magnitude as the
original vector but the opposite direction. The negative of vector quantity ⃗
A is denoted as −⃗
A
and we use a boldface minus sign to emphasize the vector nature of the quantities.

When a vector ⃗ A has magnitude |c|⃗

A is multiplied by a scalar c , the result c ⃗ A (the
absolute value of c multiplied by the magnitude of the vector ⃗
A ). If c is positive, c ⃗
A is in the
same direction as ⃗
A if c is negative, is in the direction opposite to ⃗
A.

Figure 2.2 Multiplying a vector (a) by a positive scalar and (b) by a negative scalar [1].

Vector Addition by Graphical Method

Suppose a particle undergoes a displacement ⃗

A followed by a second displacement ⃗
B.
The final result is the same as if the particle had started at the same initial point and undergone

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a single displacement ⃗
R (Figure 2.3). We call displacement ⃗
R the vector sum, or resultant, of
displacements ⃗
A and ⃗
B and we express this relationship symbolically as

⃗
R =⃗
A +⃗
B (2-1)

There are two technique in adding vectors “graphically”. First, we have tail to tip
method; the second one is the parallelogram method.

In vector addition we usually place the tail of the second vector at the head, or tip, of the
first vector (Fig. 2.3a). If we make the displacements and in reverse order, with first and second,
the result is the same (Fig. 2.3b). Thus

Figure 2.3 Tail to tip method: adding two vectors by placing them head to tail (a),
adding them in reverse order gives the same result. (b) [1].

This shows that the order of terms in a vector sum doesn’t matter. In other words, vector
addition obeys the commutative law.

⃗
R =⃗
B+⃗
A and ⃗
A+ ⃗
B= ⃗
B+⃗
A (2-2)

Figure 2.4 shows another way to represent the vector sum: If vectors ⃗
A and ⃗
B are both
drawn with their tails at the same point, vector ⃗
C is the diagonal of a parallelogram constructed
with ⃗
A and ⃗
B as two adjacent sides.

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 Figure 2.4 Parallelogram Method: adding two vectors ⃗
A and ⃗
B by constructing a
parallelogram [1].

When we need to add more than two vectors, we may first find the vector sum of any
two, add this vectorially to the third, and so on. Figure 2.5a shows three vectors ⃗ B and ⃗
A, ⃗ C . In
Figure 2.5b we first add ⃗
A and ⃗
B to give a vector sum ⃗
D we then add vectors and by the same
process to obtain the vector sum ⃗
R.

⃗
R =( ⃗
A +⃗ C =¿ ⃗
B ) +⃗ D +⃗
C (2-3)

⃗ B =¿ ⃗
A−⃗ A+ (− ⃗
B) (2-4)

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 Figure 2.6 To construct the vector difference ⃗
A−⃗
B you can either place the tail of −⃗
B at
the head of ⃗
A or place the two vectors ⃗
A and ⃗
B head to head [1].

Vector Addition by Component Method

To define what we mean by the components of a vector we begin with a rectangular

(Cartesian) coordinate system of axes (Figure 2.7a). We then draw the vector with its tail at O ,
the origin of the coordinate system. We can represent any vector lying in the xy -plane as the
sum of a vector parallel to the x -axis and a vector parallel to the y -axis. These two vectors are
labeled ⃗
A x and ⃗
A y in Figure 2.7a; they are called the component vectors of vector and their
vector sum is equal to ⃗
A . In symbols,

⃗
A=⃗
A x+ ⃗
Ay (2-4)

Now let A x be the “magnitude of vector ⃗

A x”, let A y be the “magnitude of vector ⃗
A y”, And
let A be the “magnitude of vector ⃗
A ”. Then by applying trigonometric properties and the
Phythagorean Theorem, we have the following useful equations in the case presented in Figure
2.7a and 2.7b:

A x = A cos θ , A y =A sinθ (2-5)

A=√ A x + A y (2-6)
2 2

Ay −1 A y (2-7)
tan θ= , θ=tan
Ax Ax

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 A in terms of (a) component vectors ⃗
Figure 2.7 Representing a vector ⃗ A x and ⃗
A y and
(b) components A x and A y (which in this case are both positive) [1].

A unit vector is a vector that has a magnitude of 1, with no units. Its only purpose is to
^ ) in the
point—that is, to describe a direction in space. We will always include a caret or “hat” (❑
symbol for a unit vector to distinguish it from ordinary vectors whose magnitude may or may not
be equal to 1. In the xyz -coordinate system, we define the unit vectors i^ , ^j and k^ as the unit
vectors pointing in the directions of the + x -axis, + y -axis, and + z -axis, respectively. Therefore in
two dimensional axis,

⃗
A x = A x i^ = A cos θ i^ (2-8)

⃗
A y =A y i^ =A sin θ ^j (2-9)

⃗
A=⃗
A x+ ⃗
A y =A x i^ + A y i^ =A cos θ i^ + A sin θ ^j (2-10)

Now that we already know about components and unit vectors, and how to get the
magnitude of the original vector from the magnitude of its vector components using
Pythagorean Theorem, we are ready to do vector addition by using component method. In this
method, the vectors are first resolved into their components. If the vectors lie on the xy -plane,
they are divided into their x and y components, then we calculate the x and y components of
the resultant vector by adding the x and y components of the vectors to be added. Letting the
resultant vector as ⃗
R , then we have,

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 R x =A x + B x +C x + D x + Ex … (2-11)

R y =A y +B y +C y + D y + E y … (2-12)
⃗
R x =R x i^ and ⃗
R y =R y ^j (2-13)

⃗
R =⃗
R x +⃗
R y =R x i^ + R y ^j (2.14)

To find the magnitude of the vector sum ⃗

R , we can again use the Pythagorean theorem
in Figure 2.8 together with its direction in terms of angle θ which is measured from + x –axis, we
have these two equations,

R=√ R x + R y (2-15)
2 2

Ry (2-16)
θ=tan −1
Rx

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 Example 2.1 Adding vectors using their Figure 2.8: Three successive displacements ⃗
A,
components ⃗
B and ⃗
C and the resultant (vector sum)
displacement ⃗
R =⃗
A +⃗
B+ ⃗
C [1].
Three players on a reality TV show are
brought to the center of a large, flat field.
Each is given a meter stick, a compass, a
calculator, a shovel, and (in a different
order for each contestant) the following
three displacements:

⃗
A :72.4 m ,32.0 ° east of south
⃗
B :57.3 m , 36.0° south of west
⃗
C : 17.8 . m ,due south

The three displacements lead to the point

in the field where the keys to a new
Porsche are buried. Two players start
measuring immediately, but the winner
first calculates where to go. What does
she calculate?
SOLUTION:

Execute: The angles of the vectors, measured from the + x -axis toward the + y -axis, are
( 90.0 °−32.0 ° )=58.0 ° , ( 180.0 °−36.0 ° )=216.0 ° , and 270 ° respectively. We may now use
Eq. (2-.5) to find the components of ⃗
A:
A x = A cos θ A =( 72.4 m ) ( cos 58.0 ° )=38.37 m
A y =A sinθ A =( 72.4 m )( sin 58.0 ° )=61.40 m

Distance Angle x-component y-component

A=72.4 m 58.0 ° 38.37 m 61.40 m
B=57.3 m 216.0 ° −46.36 m −33.68 m
C=17.8 m 270 ° 0.00 m −17.80 m

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 R x =−7.99 m R y =9.92 m

R=√ (−7.99 m ) + ( 9.92 m ) =12.7 m

2 2

9.92 m
θ=arctan =−51°
−7.99 m

Comparing to Fig. 2.10 shows that the calculated angle is clearly off by 180°. The correct
value is θ=180° −51° =129° , or 39 ° west of north.∎

Example 2.2 Using unit vectors

SOLUTION:
Given the two displacements

^
D=( 6.00 i+3.00
⃗ ^j−1.00 k^ ) m
and
^
E =( 4.00 i−5.00
⃗ ^j + 8.00 k^ ) m
find the magnitude of the displacement vector
⃗
R =2 ⃗
D −⃗
E.
Execute: We have
find the magnitude of the displacement vector
⃗
R =2 ⃗
D −⃗
E. ^
R =2 ( 6.00 i+3.00
⃗ ^j−1.00 k^ ) m−( 4.00 i−5.00
^ ^j+ 8.00 k^ ) m

^
R =( 12.00 i+6.00
⃗ ^j−2.00 k^ ) m−( 4.00 i−5.00
^ ^j+8.00 k^ ) m
^
R =( 12.00 i+6.00
⃗ ^j−2.00 k^ ) m−( 4.00 i−5.00
^ ^j−8.00 k^ ) m
^ ( 6.00+5.00 ) ^j+ (−2.00−8.00 ) k^ ) m
R =( ( 12.00−4.00 ) i+
⃗

^ (11.00 ) ^j+ (−10.00 ) k^ ) m

R =( ( 8.00 ) i+
⃗

The magnitude of is,

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 R=√( 8.00 m ) + ( 11.00 m ) + (−10.00 m ) =16.9 m∎
2 2 2

MOTION IN A STRAIGHT LINE

Average velocity
In this chapter we concentrate on the simplest kind of motion: a body moving along a
straight line. To describe this motion, we introduce the physical quantities velocity and
acceleration. Suppose a drag racer drives her AA-fuel dragster along a straight track (Figure.
2.9).To study the dragster’s motion, we need a coordinate system. We choose the x -axis to lie
along the dragster’s straight-line path, with the origin O at the starting line. We also choose a
point on the dragster, such as its front end, and represent the entire dragster by that point.
Hence we treat the dragster as a particle.
Let’s generalize the concept of average velocity. At time t 1 the dragster is at point P1
with coordinate x 1 with respect to the origin and at time t 2 it is at point P2 with coordinate x 2
from the origin. We define the dragsters average velocity, a vector quantity, as the rate of
change in position. That is

x 1−x 2 ∆ x (2-17)
v av− x = =
t 2−t 1 ∆t
(average x-velocity, straight-line motion)

Figure 2.9 Positions of a dragster at two times during its run [1].

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 Figure 2.10 Positions of an official’s truck at two times during its motion. The points P1
and P2 now indicate the positions of the truck, and so are the reverse of Fig. 2.8 [1].

Instantaneous velocity
The average velocity of a particle during a time interval can’t tell us how fast, or in what
direction, the particle was moving at any given time during the interval. To do this we need to
know the instantaneous velocity, or the velocity at a specific “instant” of time or specific point
along the path.
In the language of calculus, the limit of as approaches zero is called the derivative of x
with respect to t and is written dx /dt . The instantaneous velocity is the limit of the average
velocity as the time interval approaches zero; it equals the instantaneous rate of change of
position with time. We use the symbol v x with no “av ” subscript, for the instantaneous velocity
along the x -axis, or the instantaneous x -velocity:

∆ x dx (2-18)
v x = lim =
∆ t →0 ∆ t dt

(instantaneous x-velocity, straight-line motion)

Example 2.3 Average and instantaneous velocities

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 A cheetah is crouched 20 m to the east of an observer (Fig. 2.11). At time t=0 the cheetah
begins to run due east toward an antelope that is 50 m to the east of the observer. During the
first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation

x=20 m+ ( 5.0 m/s ) t . (a) Find the cheetah’s displacement between t 1=1.0 s and t 2=2.0 s (b)
2 2

Find its average velocity during that interval. (c) Find its instantaneous velocity at t 1=1.0 s by
taking ∆ t=0.1 s. (d) Derive an expression for the cheetah’s instantaneous velocity as a
function of time, and use it to find v x at t 1=1.0 s and t 2=2.0 s .
Figure 2.11 A cheetah attacking an antelope from ambush. The animals are not drawn to the

same scale as the axis [1].

(a) Find the cheetah’s displacement between SOLUTION:

t 1=1.0 s and t 2=2.0 s Execute: At t 1=1.0 s and t 2=2.0 s the
cheetah’s positions x 1 and x 2 are

x 1=20 m+ ( 5.0 m/ s 2) ( 1.0 s ) =25 m

2

x 2=20 m+ ( 5.0 m/s ) ( 2.0 s ) =40 m

2 2

The displacement during this 1.0 s

interval is

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 ∆ x=x 2−x 1=40 m−25 m=15 m
(b) Find its average velocity during that interval.
x 2−x 1 40 m−25 m
v av− x = =
t 2−t 1 2.0 s−1.0 s
v av− x =15 m/s

(c) Find its instantaneous velocity at t 1=1.0 s With the time interval ∆ t=0.1 s is from
by taking ∆ t =0.1 s. t 1=1.0 s to a new t 2=1.1 s. At t 2 the
position is

x 2=20 m+ ( 5.0 m/s 2) ( 1.1 s ) =26.05 m

2

x 2−x 1 26.05 m−25 m

v av− x = =
t 2−t 1 2.0 s−1.0 s
v av− x =10.5 m/s

(d) Derive an expression for the cheetah’s

instantaneous velocity as a function of time, dx
=( 5.0 m/s ) 2 t
2
v x=
and use it to find v x at t 1=1.0 s and t 2=2.0 s dt

.
at t 1=1.0 s

v x =( 5.0 m/s ) 2 ( 1.0 s )=10 m/ s

2

at t 1=2.0 s

v x =( 5.0 m/s 2 ) 2 ( 2.0 s )=20 m/ s ∎

24 | P a g e
Average Acceleration
Let’s consider again a particle moving along the x-axis. Suppose that at time t 1 the
particle is at point P1 and has x -component of (instantaneous) velocity v1 x, and at a later time t 2
it is at point P2 and has x -component of velocity v 2 x . So the x -component of velocity changes
∆ v x =v 2 x −v 1 x by an amount during the time interval ∆ t=t 2−t 1.
We define the average acceleration of the particle as it moves from P1 to P2 to be a
vector quantity whose x -component a av− x (called the average x-acceleration) equals ∆ v x , the
change in the x -component of velocity, divided by the time interval ∆ t :

v 2 x −v 1 x ∆ v x (2-19)
a av− x = =
t 2−t 1 ∆t

(average x-acceleration, straight-line motion)

Instantaneous acceleration
We can now define instantaneous acceleration following the same procedure
that we used to define instantaneous velocity. The instantaneous acceleration is the limit of
the average acceleration as the time interval approaches zero. In the language of calculus,
instantaneous acceleration equals the derivative of velocity with time. Thus

∆ v x dv x (2-20)
a x = lim =
∆ t →0 ∆t dt

(instantaneous x-acceleration, straight-line motion)

Example 2.4 Average and instantaneous accelerations

Suppose the x -velocity v x of the car in Fig. 2.12 at any time t is given by the equation

v x =60 m/ s + ( 0.50 m/ s 3 ) t 2

25 | P a g e
 (a) Find the change in x -velocity of the car in the time interval t 1=1.0 s to t 2=3.0 s (b) Find
the average x-acceleration in this time interval. (c) Find the instantaneous x -acceleration at
time t 1=1.0 s by ∆ t taking to be first 0.1 s , then 0.01 s , then 0.001 s . (d) Derive an expression
for the instantaneous x-acceleration as a function of time, and use it to find a x at t 1=1.0 s and
t 2=3.0 s .

Figure 2.12 A Grand Prix car at two points on the straightaway [1].

(a) Find the change in x -velocity of the SOLUTION:

car in the time interval t 1=1.0 s to Execute: Before we can apply Eq. (2-19), we

t 2=3.0 s must find the x−¿ velocity at each time from

the given equation. At t 1=1.0 s and t 2=3.0 s ,
the velocities are

v1 x =60 m/ s+ ( 0.50 m/s 3 ) ( 1.0 s ) =60.5 m/ s

2

v 2 x =60 m/s+ ( 0.50 m/s 3 ) ( 3.0 s ) =64.5m/ s

2

The change in x -velocity of the car in the

time interval t 1=1.0 s to t 2=3.0 s

∆ v x =v 2 x −v 1 x =4.0 m/ s

(b) Find the average x-acceleration in this v 2 x −v 1 x ∆ v x

a av− x = =
time interval. t 2−t 1 ∆t
∆vx 4.0 m/s 2
a av− x = = =2.0 m/s
∆ t 3.0 s−1.0 s

26 | P a g e
 (c) Find the instantaneous x -acceleration With the time interval ∆ t=0.1 s is from
at time t 1=1.0 s by ∆ t taking to be t 1=1.0 s to a new t 2=1.1 s. At t 2 proceeding
first 0.1 s . as before, we find

v 2 x =60 m/s+ ( 0.50 m/ s 3 ) ( 1.1 s ) =60.605 m/ s

2

∆ v x =v 2 x −v 1 x =60.605 m/ s−60.5 m/s=0.105m/ s

∆ v x 0.105 m/s 2
a av− x = = =1.05 m/s
∆ t 1.1 s−1.0 s

(d) Derive an expression for the By Eq. (2.20) the instantaneous x -

instantaneous x-acceleration as a dv x
=( 0.50m/ s ) 2 t
3
acceleration is a x =
function of time, and use it to find a x dt
at t 1=1.0 s and t 2=3.0 s . at t 1=1.0 s

dv x
=( 0.50m/ s ) 2 t=1.0 m/ s
3 2
ax=
dt
At t 2=3.0 s

dv x
=( 0.50m/ s ) 2 t=3.0 m/ s ∎
3 2
ax=
dt

LINEAR MOTION WITH CONSTANT ACCELERTATION

In this section, we assume that the motion is along one of the axes of the xyz -coordinate
system, say along the x -axis. As the direction of motion is specified, we can critically analyse
the motion by simply using magnitudes of related vector quantities. Note that quantities with
subscript zero “0” represent initial values of these quantities while the quantities with no
subscript represent their final values. If an object is moving with constant acceleration, then
a av− x =a x . From equation (2-19), we get

27 | P a g e
 v x =v 0 x +a x t (2-21)

by letting v1 x =v 0 x , v 2 x =v x and t 1=t 0=0 and let t 2 be any

later time t .

The average velocity is given by v av− x =x−x 0 /t−t 0 from the equation (2-17), but it can
also be expressed as v av− x =(v 0 x + v x )/ 2. Equating we have,

1 (2-22)
x−x 0= (v 0 x + v x )t
2

Substituting the expression for v x given the equation (2-21) onto equation (2-22), we have
another expression for x−x 0 as

1
x−x 0= (v 0 x + ( v 0 x +a x t ) )t ,
2
1 2 (2-23)
x−x 0=v 0 x t+ a x t ,
2

( v x −v 0 x )
Lastly, from the equation (2-21), the time t is given by t= . Substitution this into
a
equation 2-23 and simplifying, we get

x−x 0=v 0 x ( ( v x −v 0 x )
a ) (
1 v −v 2
+ ax x 0 x ,
2 ax )
v 0 x v x −v 0 x2 1
x−x 0= + (v x 2−2 v x v 0 x + v 0 x 2)
ax 2 ax
2 2 2
2 a x ( x−x 0)=2 v 0 x v x −2 v 0 x +v x −2 v x v 0 x + v 0 x
2 a x ( x−x 0)=v x 2−v 0 x 2
(2-24)
v x 2−v 0 x 2=2 a x ( x− x0 )

28 | P a g e
 Equations 2-21 to 2-24 described a motion along the straight line with constant
acceleration. If we assume that a “positive” directions coincides with the direction of the +x-axis,
then x and x 0 are positive if measured from the origin to the right. “Positive” values v x and v 0 x
meant that the motion is going to the right while “negative” values would mean that the motion is
going to the left. However, “positive” acceleration means that the object is going faster, that is v x
> v 0 x ; ant it is “negative” if it is decelerating or slowing down, v x < v 0 x . Table 2.1 summarizes the
equations of linear motion with constant acceleration.
Table 2.1: Equation of linear motion with constant acceleration

x x0 vx v0x ax t
v x =v 0 x +a x t x x x x
1 x x x x x
x−x 0= (v 0 x + v x )t
2
1 2 x x x x x
x−x 0=v 0 x t+ a x t
2
2 2
v x −v 0 x =2 a x (x− x0 ) x x x x x

Example 2.5: Constant-acceleration calculations

A motorcyclist heading east through a small town accelerates at a 4.0 m/ s2 constant after he
leaves the city limits (Fig. 2.13). At time t=0 s he is 5.0 m east of the city-limits signpost,
moving east at 15 m/s (a) Find his position and velocity at t=2.0 s (b) Where is he when his
velocity is 25 m/s ?

29 | P a g e
 Figure 2.13 A motorcyclist traveling with constant acceleration [1].

SOLUTION:
(a) Find his position and velocity at Execute: Eq. (2.23) and the x-velocity at this
t=2.0 s time by using Eq. (2.21):

Eq. (2.23)
1
x=x 0 +v 0 x t + a x t 2
2
1
x=5.0 m+ ( 15 m/s )( 2.0 s ) + ( 4.0 m/ s ) ( 2.0 s ) =43 m
2 2
2

Eq. (2.21)
v x =v 0 x +a x t

v x =( 15 m/s ) + ( 4.0 m/s 2 ) ( 2.0 s )=23 m/s

(b) Where is he when his velocity is We want to find the value of x when
25 m/s ? v x =25 m/s but we don’t know the time when
the motorcycle has this velocity.
So we use Eq. (2-24)

2 2
v x −v 0 x =2 a x (x− x0 )

Solving for x ,

2 2
v x −v 0 x
x=x 0 +
2 ax

30 | P a g e
 ( 25 m/ s )2−( 15 m/s )2
x=5.0 m+ =55 m∎
2 ( 4.0 m/ s2 )

FREE FALL
An example of a linear motion with constant acceleration is free fall. Free fall is the
motion of a falling body under the influence of the earth’s gravity. The constant acceleration of
this body is called the acceleration due to gravity, denoted by g is approximately −9.8 m/s 2
near the earth’s surface. The negative sign indicates that g is directed downward. Here we will
denote the value of a in equations 2-21 to 2-23 as g and consider only vertical motion of a body.

Example 2.6 A freely falling coin

A one-euro coin is dropped from the Leaning Tower of Pisa and falls freely from rest. What
are its position and velocity after 1.0 s ,2.0 s , and 3.0 s ?
Figure 2.14 A coin freely falling from rest [1].

31 | P a g e
 What are its position and velocity after SOLUTION:
1.0 s ,2.0 s , and 3.0 s ? Execute: Using Kinematics equation along y-
axis we can find the expression of position:
1
y− y 0=v 0 y t+ a y t 2
2

1 2
y= y 0+ v 0 y t+ a y t
2

1 2
y=0+0+ (−g ) t
2
1
y= (−9.8 m/s ) t =(−4.9 m/s ) t
2 2 2 2
2

And velocity:
v y =v 0 y + a y t

v y =0+ (−9.8 m/ s2 ) t

v y =(−9.8 m/ s2 ) t
When 1.0 s ,2.0 s , and 3.0 s ?
The position and velocity is given by:

When 1.0 s , y=−4.9 m, v y =−9.8 m/s ,

When 2.0 s , y=−20 m, v y =−20 m/s ,

When 3.0 s , y=−44 m , v y =−29 m/s ∎

32 | P a g e
B. Performance Task:
I. Problem solving: show your solutions completely and neatly.
1. A motorist traveling with a constant speed 15 m/s of about passes a school-crossing
corner, where the speed limit is about 10 m/s. Just as the motorist passes the school
crossing sign, a police officer on a motorcycle stopped there starts in pursuit with a
constant acceleration of 3.0 m/s2 (Fig. 2.15). (a) How much time elapses before the
officer passes the motorist? (b) What is the officer’s speed at that time? (c) At that time,
what distance has each vehicle travelled?
2. You throw a ball vertically upward from the roof of a tall building. The ball leaves your
hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is
then in free fall. On its way back down, it just misses the railing. Find (a) the ball’s
position and velocity 1.00 s and 4.00 s after leaving your hand; (b) the ball’s velocity
when it is 5.00 m above the railing; (c) the maximum height reached; (d) the ball’s
acceleration when it is at its maximum height.
3. At what time after being released has the ball in Problem # 2 fallen 5.00 m below the
roof railing?
4. A Tennis Serve. In the fastest measured tennis serve, the ball left the racquet at 73.14
m/s. A served tennis ball is typically in contact with the racquet for 30.0 ms and starts
from rest. Assume constant acceleration. (a) What was the ball’s acceleration during this
serve? (b) How far did the ball travel during the serve?

33 | P a g e
 5. BIO Automobile Airbags. The human body can survive an acceleration trauma incident
(sudden stop) if the magnitude of the acceleration is less than 250 m/s2. If you are in an
automobile accident with an initial speed of 105 km/h and you are stopped by an airbag
that inflates from the dashboard, over what distance must the airbag stop you for you to
survive the crash?

Figure 2.15 a) Motion with constant acceleration overtaking motion with constant
velocity. (b) A graph of x versus t for each vehicle.

34 | P a g e
Exercise – MODULE 2
Name:__________________________________________________________Section:_____
II. Multiple Choice: Write the letter of your choice on the space provided. Then Explain or
show the solution that justifies your answer. Each item is worth 2 points; correct answers
but without solution will gain only 1 point.

_____1. A car travels in the + x -direction on a straight and level road. For the first 4.00 s
of its motion, the average velocity of the car is v av− x =6.25 m/s . How far does the car
travel in 4.00 s ?
a) 25.0 m b) 22.0 m c) 25.0 m/s d) 25 s

_____2. A bird is flying due east. Its distance from a tall building is given by
x ( t )=28.0 m+ ( 12.4 m/s ) t−( 0.0450 m/s ) t . What is the instantaneous velocity of the bird
3 3

when t=8.00 s.

a) 7.36 m/s b) 3.76 m/s c) 3.67 m/s d) 6.73 m/s

_____3. A race car starts from rest and travels east along a straight and level track. For
the first 5.0 s of the car’s motion, the eastward component of the car’s velocity is given by
v x ( t )=( 0.860 m/ s3 ) t 2. What is the acceleration of the car when v x =16.0 m/s .
a) 7.24 m/s2 b) 4.27 m/s2 c) 2.42 m/s2 d) 7.42 m/s2

_____4. An antelope moving with constant acceleration covers the distance between
two points 70.0 m apart in 7.00 s. Its speed as it passes the second point is 15.0 m/s.
What is its speed at the first point?
a) 5.0 m/s b) 5.2 m/s c) 4.0 m/s d) 4.2 m/s

_____5. Referring to problem # 4: What is its acceleration?

a) 1.43 m/s b) 1.43 m/s2 c) 1.34 m/s2 d) 1.34 m/s

35 | P a g e
 _____6. The fastest measured pitched baseball left the pitcher’s hand at a speed of
45.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and
produced constant acceleration, what acceleration did he give the ball?
a) 625 m/s2 b) 645 m/s2 c)675 m/s2 d) 655 m/s2

_____7. Referring to problem # 6: how much time did it take him to pitch it?
a) 0.00667 s b) 0.667 s c) 0.0667 s d) 0.000667 s

_____8. If a flea can jump straight up to a height of 0.440 m, what is its initial speed as
it leaves the ground?
a) 2.94 m/s b) 94.2 m/s c) 4.92 m/s d) 2.24 m/s

_____9. Referring to problem # 8: How long is it in the air?

a) 0.600 s b) 0.060 s c) 0.006 s d) 0.000600 s

_____10. A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s.
How much time elapses until the bowling pin returns to the juggler’s hand?
a) 1.77 s b) 1.67 s c) 1.57 s d) 1.47 s

36 | P a g e
Summary

Scalars, vectors, and vector addition:

Scalar quantities are numbers and combine with the usual rules of arithmetic. Vector
quantities have direction as well as magnitude and combine according to the rules of vector
addition. The negative of a vector has the same magnitude but points in the opposite
direction.

Vector components and vector addition:

Vector addition can be carried out using components of vectors. The x -component of
⃗
R =⃗
A +⃗
B is the sum of the x -components of and and likewise for the y - and z -components.

Unit vectors:
Unit vectors describe directions in space. A unit vector has a magnitude of 1, with no units.
The unit vectors i^ , ^j and k^ aligned with the x -, y -, and z -axes of a rectangular coordinate
system, are especially useful.

Straight-line motion, average and instantaneous x-velocity:

When a particle moves along a straight line, we describe its position with respect to an origin
O by means of a coordinate such as x . The particle’s average x-velocity v av− x during a time
interval ∆ t=t 2−t 1 is equal to its displacement ∆ x=x 2−x 1 divided by ∆ t . The instantaneous
x-velocity v x at any time t is equal to the average x-velocity for the time interval from t to t+ ∆ t
in the limit that ∆ t goes to zero. Equivalently, v x is the derivative of the position function with
respect to time.

Average and instantaneous x-acceleration:

The average x-acceleration a av− x during a time interval ∆ t is equal to the change in velocity
∆ v x =v 2 x −v 1 x during that time interval divided by ∆ t . The instantaneous x -acceleration is the
limit of as goes to zero, or the derivative of v x with respect to t .

Straight-line motion with constant acceleration:

When the x-acceleration is constant, four equations relate the position x and the x-velocity v x
at any time t to the initial position x 0 the initial x-velocity v 0 x (both measured at time and the

37 | P a g e
 x-acceleration a x .

Freely falling bodies: Free fall is a case of motion with constant acceleration. The
magnitude of the acceleration due to gravity is a positive quantity, g. The acceleration of a
body in free fall is always downward.

Post Test: This post-test can be conducted through online or face to face.

References
2. Kramer, Laird.Young, Hugh D. (2012) Study guide, Sears & Zemansky's University
physics, 13th edition, Young and Freedman /San Francisco, CA : Pearson.

38 | P a g e
 MODULE 3

Motion in two or three dimensions

Time frame:

Objective: By studying this chapter you will learn how tp:

 Represent the position of a body in two or three dimensions using vectors.

 Determine the vector velocity of a body from a knowledge of its path.
 Find the vector acceleration of a body, and why a body can have an acceleration even if
its speed is constant.
 Interpret the components of a body’s acceleration parallel to and perpendicular to its
path.
 Describe the curved path followed by a projectile.
 The key ideas behind motion in a circular path, with either constant speed or varying
speed.

Introduction:

We can’t answer these kinds of questions using the techniques of Module 2, in which
particles moved only along a straight line. Instead, we need to extend our descriptions of motion
to two- and three-dimensional situations. We’ll still use the vector quantities displacement,
velocity, and acceleration, but now these quantities will no longer lie along a single line. We’ll
find that several important kinds of motion take place in two dimensions only—that is, in a
plane. We can describe these motions with two components of position, velocity, and
acceleration.

Pre-test: This pre-test will be conducted through online or face to face oral recitation.

39 | P a g e
Learning Activities

A. Content/Discussion

AVERAGE AND INSTANTANEOUS VELOCITY

Position Vector
To describe the motion of a particle in space, we must first be able to describe the
particle’s position. Consider a particle that is at a point P at a certain instant. The position
vector r⃗ of the particle at this instant is a vector that goes from the origin of the coordinate
system to the point P in Figure 3.1. The Cartesian coordinates x , y , and z of point P are the x -,
y -, and z -components of vector Using the unit vectors we introduced in Section 1.9, we can
write

^ y ^j + z k^
r⃗ =x i+ (3-1)
(Position vector)

Figure 3.1 The position vector from the origin to point P has components x , y , and z .
The path that the particle follows through space is in general a curve (Fig. 3.2)

40 | P a g e
 Figure 3.2 The average velocity ⃗v av between points P1 and P2 has the same direction as
the displacement ∆ r⃗ .

Velocity Vector
During a time interval ∆ t the particle moves from P1, where its position vector is r⃗ 1, to
P2 where its position vector is r⃗ 2 . The change in position (the displacement) during this interval

∆ r⃗ =r⃗ 2− ⃗r 1=( x 2−x 1 ) i^ + ( y 2− y 1) ^j+ ( z2 −z 1) k^ . We define the average velocity ⃗v av during this
interval in the same way we did in Chapter 2 for straight-line motion, as the displacement
divided by the time interval:

r⃗ 2 −⃗r 1 ∆ ⃗r (3-2)
⃗v av = =
t 2 −t 1 ∆ t
(average velocity vector)

Instantaneous velocity
We now define instantaneous velocity just as we did in Chapter 2: It is the limit of the
average velocity as the time interval approaches zero, and it equals the instantaneous rate of
change of position with time. The key difference is that position r⃗ and instantaneous velocity ⃗v
are now both vectors:

(3-3)

41 | P a g e
 ∆ r⃗ d ⃗r
⃗v = lim =
∆ t →0 ∆ t dt
(instantaneous velocity vector)

It follows that the components and of the instantaneous velocity are simply the time derivatives
of the coordinates x , y ,and z . That is,

d ^
⃗v = ( x i + y ^j + z k^ )
dt

dx ^ d y ^ dz ^
⃗v = i+ j+ k
dt dt dt

^ y ^j+ v z k^
⃗v =v x i+v

dx dy dz
v x= v y= v z=
dt dt dt (3-4)

(components of
instantaneous velocity)

The magnitude of the instantaneous velocity vector ⃗v —that is, the speed—is given in
terms of the components v x , v y and v z by the Pythagorean relation:

|⃗v|=v=√ v x 2+ v y 2+ v z2 (3-4)

When the particle moves in the xy−¿ plane, In this case, z and v z are zero. Then the
speed (the magnitude of ⃗v is

|⃗v|=v=√ v x 2+ v y 2 (3-5)

and the direction of the instantaneous velocity is given by the angle α (the Greek letter alpha) in
the Figure 3.3. We see that

42 | P a g e
 vy (3-6)
tan α=
vx

Figure 3.3 The two velocity components for motion in the xy -plane [1].

Example 3.1 Calculating average and Figure 3.4 At t=0.0 s the rover has position
instantaneous velocity vector r⃗ 0 and instantaneous velocity vector ⃗v 0
. Likewise, r⃗ 1 and ⃗v1 are the vectors at
A robotic vehicle, or rover, is exploring the
t=1.0 s ; r⃗ 2 and r⃗ 2 are the vectors t=2.0 s [1].
surface of Mars. The stationary Mars lander
is the origin of coordinates, and the
surrounding Martian surface lies in the xy -
plane. The rover, which we represent as a
point, has x - and y -coordinates that vary with
time:

x=2.0 m−( 0.25 m/ s ) t

2 2

y= (1.0 m/ s ) t + ( 0.025 m/ s ) t
3 3

a) Find the rover’s coordinates and distance

from the lander at t=2.0 s (b) Find the rover’s
displacement and average velocity vectors

43 | P a g e
 for the interval t=0.0 s to t=2.0 s (c) Find a
general expression for the rover’s
instantaneous velocity vector ⃗v . Express ⃗v at
t=2.0 s in component form and in terms of
magnitude and direction.
Find the rover’s coordinates and distance SOLUTION:
from the lander at t=2.0 s Execute: The rovres coordinates and
distance from the lander at t=2.0 s

x=2.0 m−( 0.25 m/ s ) ( 2.0 s ) =1.0 m

2 2

y= (1.0 m/ s ) (2.0 s)+ ( 0.025 m/s ) ( 2.0 s ) =2.0 m

3 3

The rover’s distance from the origin at this

time is

|r⃗|=r =√ (1.0 m )2 + ( 2.0 m )2=2.4 m

(b) Find the rover’s displacement and To find the displacement and average
average velocity vectors for the interval velocity over the given time interval, we first
t=0.0 s to t=2.0 s express the position vector as a function of
time t. From Eq. (3-1) this is

^ y ^j
r⃗ =x i+
r⃗ =[ 2.0 m−( 0.25 m/s ) t ] i+ [ ( 1.0 m/ s ) t+ ( 0.025 m/s3 ) t 3 ] ^j
2 2 ^

At t=0.0 s,

^ ( 0.0 m ) ^j
r⃗0 =( 2.0 m) i+

At t=2.0 s

r⃗ 2=( 2.0 m−( 0.25 m/s ) ( 2.0 s ) ) i+

^ ( ( 1.0 m/s ) ( 2.0 s )+ ( 0.025 m/s3 ) (
2 2

44 | P a g e
 ^ (2.2 m ) ^j
r⃗ 2= (1.0 m ) i+

The displacement from t=0.0 s to t=2.0 s is

therefore
^ ( 2.2 m−0.0 m ) ^j
∆ r⃗ =⃗r 2−⃗r 0=( 1.0 m−2.0 m ) i+

^ ( 2.2 m ) ^j
∆ r⃗ =⃗r 2−⃗r 0=(−1.0 m ) i+

The average velocity over this interval is the

displacement divided by the elapsed time:

^ ( 2.2 m ) ^j
r⃗ 2 −⃗r 1 (−1.0 m ) i+
⃗v av = =
t 2 −t 1 2.0 s−0.0 s

^ ( 2.2 m ) ^j
(−1.0 m ) i+
⃗v av =
2.0 s
^ ( 1.1 m/ s ) ^j
⃗v av =(−0.50 m/s ) i+

The components of this average velocity are

⃗v av− x =−0.50 m/s and ⃗v av− y =1.1m/ s.
(c) Find a general expression for the rover’s From Eq. (3-4) the components of
instantaneous velocity vector ⃗v . Express ⃗v at instantaneous velocity are the time
t=2.0 s in component form and in terms of derivatives of the coordinates:
magnitude and direction.
^ y ^j+ v z k^
⃗v =v x i+v

dx dy dz
vx= v y= v z=
dt dt dt

dx
=(−0.25 m/s ) 2 t
2
vx=
dt
dy
=( 1.0 m/ s ) + ( 0.025m/ s ) 3 t
3 2
v y=
dt

45 | P a g e
 ⃗v =[ ( 0.25 m/s 2 ) 2t ] i+
^ [ (1.0 m/ s ) + ( 0.025 m/s 3 ) 3 t 2 ] ^j

At t=2.0 s the velocity vector v⃗ 2 has

components

v 2 x =(−0.25 m/s 2 ) 2 ( 2.0 s )=−1.0 m/ s

v 2 y =( 1.0 m/ s ) + ( 0.025m/ s ) 3 ( 2.0 s ) =1.3 m/s

3 2

The magnitude of the instantaneous velocity

(that is, the speed) at t=2.0 s is

v 2=√ v 2 x 2 +v 2 y 2

v 2=√ (−1.0 m/ s ) + ( 1.3 m/s )

2 2

v 2=1.6 m/s

Figure 3.8 shows the direction of the velocity

vector ⃗v , which is at an angle α between 90 °
and 180 ° with respect to the positive
x-axis. From Eq. (3-6) we have

vy 1.3 m/ s
α =arctan =arctan =−52 °
vx −1.0 m /s
This is off by 180 °; the correct value of the
angle is α =180 °−52 °=128 ° , or 38 ° west of
north.∎

AVERAGE AND INSTATANEOUS ACCELERATION

Average Acceleration
Since we now treat velocity as a vector, acceleration will describe changes in the
velocity magnitude (that is, the speed) and changes in the direction of velocity (that is, the
direction in which the particle is moving). In Figure 3.5, a car (treated as a particle) is moving

46 | P a g e
along a curved road. The vectors ⃗v1 and ⃗v 2 represent the car’s instantaneous velocities at time
t 1 when the car
is at point P1 and at time t 2 when the car is at point P2. We define the average acceleration of
the car during this time interval as the velocity change divided by the time interval ∆ t=t 2−t 1 :

v⃗ 2−⃗v 1 ∆ ⃗v (3-6)
a⃗ av = =
t 2−t 1 ∆t

Figure 3.5 (a) A car moving along a curved road from P1 to P2 (b) How to obtain the
∆ ⃗v
change in velocity ∆ ⃗v =⃗v 2−⃗v 1 by vector subtraction. (c) The vector a⃗ av = represents
∆t
the average acceleration between P1 and P2.

Instantaneous Acceleration
As in Chapter 2, we define the instantaneous acceleration (a vector quantity) at point P1 as
the limit of the average acceleration vector when point approaches point P2, so and both
approach zero. The instantaneous acceleration is also equal to the instantaneous rate of
change of velocity with time:

∆ ⃗v d ⃗v (3-7)
a⃗ = lim =
∆ t →0 ∆ t dt

47 | P a g e
Each component of the acceleration vector is the derivative of the corresponding component of
velocity:

d ^
a⃗ = ( v i^ + v y ^j+ v z k)
dt x

d vx d v y d v z
a⃗ = ^
i+ ^j+ k^
dt dt dt

^ y ^j+ a z k^
a⃗ =a x i+a

d vx d vy d vz (3-8)
ax= a y= az =
dt dt dt
Or
d2 x ^ d2 y ^ d2 z ^
a⃗ = i+ j+ k
dt dt dt
(3-9)
d2 x d2 y d2 z
ax= a y= az =
dt dt dt

(components of
instantaneous acceleration)

When the particle moves in the xy−¿ plane, In this case, z and a z are zero. Then the
acceleration (the magnitude of a⃗ ) is

|⃗a|=a=√ v x 2+ v y 2 (3-10)

and the direction of the instantaneous acceleration is given by the angle α (the Greek letter
alpha) that is

48 | P a g e
 ay (3-11)
tan α=
ax

Example 3.2 Calculating average and Figure 3.6 The path of the robotic rover,
instantaneous acceleration showing the velocity and acceleration at and
and t=0.0 s ( ⃗v 0 , a⃗ 0 ), t=1.0 s ( ⃗v 1 , ⃗a1 ) , and
Let’s return to the motions of the Mars rover
t=2.0 s ( ⃗v 2 , ⃗a2 ) .
in Example 3.1. (a) Find the components of
the average acceleration for the interval
t=0.0 s to t=2.0 s (b) Find the instantaneous
acceleration at t=2.0 s .

In Example 3.1 we found the components

of the rover’s instantaneous velocity at any
timet :
dx
v x= =(−0.25 m/s 2 ) 2 t=(−0.50 m/ s2 ) t
dt

dy
v y= =( 1.0 m/ s ) + ( 0.025m/ s3 ) 3 t 2
dt

¿ ( 1.0 m/s )+ ( 0.075 m/s ) t

3 2

49 | P a g e
 (a) Find the components of the average In Example 3.1 we found that at t=0.0 s the
acceleration for the interval t=0.0 s to t=2.0 s velocity components are

v x =0.0 m/s , v y =1.0 m/s

at t=1.0 s

v x =−1.0 m/s , v y =1.3 m/s

Thus the components of average

acceleration in the interval t=0.0 s to t=2.0 s
∆ v x −1.0 m/s−0.0 m/ s 2
a av− x = = =−0.50 m/ s
∆t 2 .0 s−0.0 s

∆ y 1.3 m/s−1.0 m/s

a av− y = =
∆t 2 .0 s−0.0 s
2
¿ 0.15 m/s

(b) Find the instantaneous acceleration at d vx

a x= =−0.50m/ s2
t=2.0 s . dt

d vy
=( 0.075 m/s ) 2 t
3
a y=
dt

Hence the instantaneous acceleration a⃗

vector at time t is

2 ^
a⃗ =(−0.50 m/ s ) i+ ( 0.15 m/s3 t ) ^j

at t=2.0 s the components of acceleration

and the acceleration vector are

50 | P a g e
 d vx 2
a x= =−0.50m/ s
dt

d vy 2
a y= =0.30 m/s
dt

2 ^
a⃗ =(−0.50 m/ s ) i+ ( 0.30 m/s3 t ) ^j

The magnitude of acceleration at this time is

a=√ a x 2+ a y 2

√ 2 2
a= (−0.50 m/s 2 ) + ( 0.30 m/ s 2 ) =0.58 m/s 2

A sketch of this vector (Fig. 3.6) shows that

the direction β angle of with respect to the
positive x -axis is between 90 ° and 180 °.
From Eq. (3-11) we have

ay 0.30 m/s
2
α =arctan =arctan =31 °
ax −0.50 m/s 2
β=180 ° + (−31 ° )=149 ° ∎

PROJECTILE MOTION
A projectile is any body that is given an initial velocity and then follows a path
determined entirely by the effects of gravitational acceleration and air resistance. A batted
baseball, a thrown football, a package dropped from an airplane, and a bullet shot from a rifle
are all projectiles. The path followed by a projectile is called its trajectory. The key to analyzing
projectile motion is that we can treat the x−¿ and y−¿coordinates separately. For the
acceleration the x−¿ component of acceleration is zero, and the y−¿component is constant
and equal to – g . Figure 3.7 shows the trajectory of a projectile that starts at (or passes through)
the origin at time t=0 , along with its position, velocity, and velocity components at equal time
intervals. The x -component of acceleration is zero, so v x is constant. The y -component of

51 | P a g e
acceleration is constant and not zero, so v y changes by equal amounts in equal times, just the
same as if the projectile were launched vertically with the same initial y -velocity.

Figure 3.7 The trajectory of an idealized [1].

Figure 3.8 If air resistance is negligible, the trajectory of a projectile is a combination of

horizontal motion with constant velocity and vertical motion with constant acceleration

[1].

52 | P a g e
 We now work through several examples of projectile motion qualitatively. We use the
kinetic equations separately for the vertical and horizontal components of the motion. These
equations are shown separately for the x and y components of the motion in Table 3.1, for the
general case of two dimensional motion.

Table 3.1 General Kinematic Equation for Constant Acceleration in Two Dimensions

x -component y -component
v x =v 0 x +a x t v y =v 0 y + a y t
1 1
x−x 0= (v 0 x + v x )t y− y 0= ( v 0 y + v y )t
2 2
1 2 1 2
x−x 0=v 0 x t+ a x t y− y 0=v 0 y t+ a y t
2 2
2 2 2 2
v x −v 0 x =2 a x (x− x0 ) v y −v 0 y =2 a y ( y− y 0 )

IF we simplify these equations for the case of projectile motion because we can set a x =0 , and
a y =−g . Table 3.2 shows the equations applicable to critical analyze projectile motion. Take
note that free fall, vertical displacements above the origin is considered positive.

Table 3.2 Kinematic Equation for Projectile Motion

x -component y -component
v x =v 0 x v y =v 0 y −¿
1
y− y 0= ( v 0 y + v y )t
2
x−x 0=v 0 x t 1
y− y 0=v 0 y t− g t 2
2
2 2
v y −v 0 y =−2 g( y − y 0 )

We can also represent the initial velocity ⃗v 0 by its magnitude (the initial speed) and its angle with
the positive x -axis shown in Figure 3.9.

53 | P a g e
 Figure 3.9 The initial velocity components v 0 x and v 0 y of a projectile (such as a kicked

soccer ball) are related to the initial speed v 0 and initial angle α 0.

In terms of these quantities, the components v 0 x and v 0 z of the initial velocity are

v 0 x =v 0 cos α 0 , v 0 y =v 0 sin α 0 (3-12)

If we substitute these relationships in kinematics equations and set x 0= y 0 =0, we find

v x =v 0 cos α 0 (3-13)
x=v 0 cos α 0 t (3-14)

v y =v 0 sin α 0−¿ (3-15)

1 2 (3-16)
y=v 0 sin α 0 t− g t
2

We can get a lot of information from Eqs. (3-13) through (3.16). For example, at any time the
distance r of the projectile from the origin (the magnitude of the position vector r⃗ ) is given by

|r⃗|=r =√ x 2 + y 2 (3-17)

The projectile’s speed (the magnitude of its velocity) at any time is

|⃗v|=v=√ v x 2+ v y 2 (3-18)

The direction of the velocity, in terms of the angle α it makes with the positive x -direction see
Figgure 3.8, is given by

54 | P a g e
 vy (3-19)
tan α=
vx

The expression for the maximum height and range for an object in projectile motion are

−v 0 x 2 (3-20)
h= y=
2g
(height)

R=x=v 0 x t=v 0 x ( −2gv )= −2 vg v

0y 0x 0y
(3-19)

−2 v 02 sin α 0 cos α 0
¿
g

(Range)

Example 3.3 A body projected horizontally Figure 3.10 Our sketch for this problem.

A motorcycle stunt rider rides off the edge of

a cliff. Just at the edge his velocity is
horizontal, with magnitude 9.0 m/ s . Find the
motorcycle’s position, distance from the edge
of the cliff, and velocity 0.50 s after it leaves
the edge of the cliff.

Find the motorcycle’s position, distance from From Eq. (3-14) and (3-16), the motorcycle’s
the edge of the cliff, and velocity 0.50 s after it x - and y -coordinates at t=0.50 s are
leaves the edge of the cliff.
x=v 0 cos α 0 t=( 9.0 m/s ) cos ( 0 ° )( 0.50 s ) =4.5 m

55 | P a g e
 1 2
y=v 0 sin α 0 t− g t
2

1 2 2
y= ( 9.0 m/s ) sin ( 0 ° )( 0.50 s ) − (9.8 m/s ) ( 0.50 s )
2
1 2 2
y=0− (9.8 m/s ) ( 0.50 s )
2
−1
y= ( 9.8 m/s2 ) ( 0.50 s )2=−1.2 m
2
Thus the motorcyles position is given by
position vector r⃗ :
^ (−1.2 m ) ^j
r⃗ =( 4.5 m ) i+

The negative value of y shows that the

motorcycle is below its starting point.

From Eq. (3-17), the motorcycle’s distance

from the origin at t=−0.50 s is

r =√ x + y =√ ( 4.5 m ) + (−1.2 ) =4.7 m

2 2 2 2

From Eq. (3-13) and (3-15), the velocity

components at
t=0.50 sare

v x =v 0 cos α 0 =( 9.0 m/s ) cos ( 0° ) =9.0 m/s

v y =v 0 sin α 0 – >¿−¿=−( 9.8 m/s 2 ) ( 0.50 s )=−4.9 m/s

The motorcycle has the same horizontal

velocity v x as when it left the cliff at t=0 , but
in addition there is a downward (negative)
vertical velocity v y. The velocity vector at
t=0. 50 s is

56 | P a g e
 ^
⃗v =v x i+v ^
yj

⃗v =( 9.0 m/s ) i^ + (−4.9 m/s ) ^j

From Eq. (3-18), the speed (magnitude of

the velocity) at t=0.50 s is

v=√ v x2 + v y 2

v=√ ( 9.0 m/s ) + (−4.9 m/s ) =10.2m/ s

2 2

From Eq. (3.26), the angle of the velocity

vector is

( )
vy −4.9 m/s
α =arctan =arctan =−29 °
vx 9.0 m/ s
The velocity 29 ° is below the horizontal0.∎

Example 3.7 Height and range of a Figure 3.11 Our sketch for this problem.
projectile I: A batted baseball

A batter hits a baseball so that it leaves the

bat at speed v 0=37.0 m/s at an angle
α 0=53.1 °. (a) Find the position of the ball
and its velocity (magnitude and direction) at
t=2. 00 s . (b) Find the time when the ball
reaches the highest point of its flight, and its
height h at this time. (c) Find the horizontal
range R —that is, the horizontal distance from
the starting point to where the ball hits the
ground.
(a) Find the position of the ball and its SOLUTION:
velocity (magnitude and direction) at Execute: (a) We want to find x , y , and v x at
t=2. 00 s . t=2.00 s . The initial velocity of the ball has

57 | P a g e
 components

v 0 x =v 0 cos α 0 =( 37.0 m/ s ) ( cos 53.1 ° )=22.2 m/s

v 0 y =v 0 cos α 0=( 37.0 m/ s ) ( sin 53.1° )=29.6 m/ s

From Eq. (3-13) through (3.16),

x=v x 0 t= ( 22.2m/s ) ( 2.00 s )=44.5 m

1 2 29.6 m
y=v y 0 t− g t =
2 s ( 1
)
( 2.00 s )− ( 9.8 m/s 2 ) ( 2.00 s )2=39.6 m
2

v x =v 0 cos α 0 =22.2m/ s

m
−( 9.8 m/ s ) ( 2.00 s )=10 m/s
2
v y =v 0 sin α 0−¿=29.6
s

The y -component of velocity is positive at

t=2.00 s , so the ball is still moving upward
(Fig. 3.23). From Eqs. (3-18) and (3-19), the
magnitude and direction of the velocity are

v=√ v x2 + v y 2

v=√ ( 22.2m/ s ) + ( 10.0 m/ s ) =24.4 m/s

2 2

( )
vy 10.0 m/s
α =arctan =arctan =24.2 °
vx 22.2 m/s

The direction of the velocity (the direction of

the ball’s motion) is 24.2 ° above the
horizontal.

(b) Find the time when the ball reaches the At the highest point, the vertical velocity v y is
highest point of its flight, and its height h at zero. Call the time when this happens t 1; then
this time.

58 | P a g e
 v y =v 0 y −g t 1
0=v 0 y −g t 1
v 0 y =g t 1

g 29.6 m/s
t 1= = =3. 02 s
v 0 y 9.80 m/s2
The height h at the highest point is the value
of y at time t 1:
1 2
h=v 0 y t 1 − g t 1
2
1
¿ ( 29.6 m/s )( 3. 02 s )− ( 9.80 m/s 2 ) ( 3.02 s )2
2
¿ 44.7 m

We’ll find the horizontal range in two steps.

First, we find the time t 2 when y=0 (the ball
is at ground level):

1
y=0=v 0 y t 2− g t 22
2

1
2
2
( 1
0=v 0 y t 2− g t 2 =t 2 v 0 y − g t 2
2 )
This is a quadratic equation for t 2. It has two
roots:
t 2=0 s
and
2 v 0 y 2 ( 9.6 m/ s )
t 2= = =6.04 s
g 9.80 m/s 2

The ball is at y=0 at both times. The ball

59 | P a g e
 leaves the ground at t=0 , and it hits the
ground at t=6.04 s.The horizontal range R is
the value of x when the ball returns to the
ground at t=6.04 s:

R=x=v 0 x t=( 22.2 m/s )( 6.04 s ) =134 m

The vertical component of velocity when the

ball hits the ground is

v y =v 0 y −g t 2=29.6
m
s ( m
)
− 9.80 2 ( 6.04 s )=−29.6 m/s
s

That is, v y has the same magnitude as the

initial vertical velocity v 0 y but the opposite
direction (down). Since v x is constant, the
angle α =53.1 ° (below the horizontal) at this
point is the negative of the initial angle
α 0=53.1 ° ∎

MOTION IN A CIRCLE (Optional)

When a particle moves in a circular path of radius R with constant speed v (uniform
circular motion), its acceleration a⃗ is directed toward the center of the circle and perpendicular
to ⃗v. The magnitude a rad of the acceleration can be expressed in terms of v and R or in terms of
2 πR
R and the period T (the time for one revolution), where v= .
T
If the speed is not constant in circular motion (nonuniform circular motion), there is still a
radial component of a⃗ given by Eq. (3-19) or (3-20), but there is also a component of a⃗ parallel
(tangential) to the path. This tangential component is equal to the rate of change of speed,
dv /dt .

60 | P a g e
 v2 (3-19)
a rad =
R
4 πR (3-20)
a rad = 2
T

Example 3.8 Centripetal acceleration on a SOLUTION

curved road

An Aston Martin V8 Vantage sports car has a

“lateral acceleration” of

0.96 g=( 0.96 ) ( 9.8 m/s )=9.4 m/ s . This is the

2 2

maximum centripetal acceleration the car can

sustain without skidding out of a curved path.
If the car is traveling at a constant 40 m/ s
(about or ) on level ground, what is the radius
R of the tightest unbanked curve it can
negotiate?
EXECUTE: The car is in uniform circular
motion because it’s moving at a constant
speed along a curve that is a segment of a
circle. Hence we can use Eq. (3-19) to solve
for the target variable R in terms of the given
centripetal acceleration
2
v
a rad =
R

2 2
v ( 40 m/s )
R= = =170 m
a rad ( 9.4 m/s 2 )

This is the minimum radius because a rad is

the maximum centripetal acceleration.∎

61 | P a g e
 B. Performance Task:

I. Problem solving: show your solutions completely and neatly.

1. You throw a ball from your window 8.0 m above the ground. When the ball leaves
your hand, it is moving at 10.0 m/s at an angle 20 ° of below the horizontal. How
far horizontally from your window will the ball hit the ground? Ignore air
resistance.
2. Find the maximum height h and horizontal range R (see Fig. 3.11) of a projectile
launched with speed v 0 at an initial angle α 0 between 0 ° and 90 ° . For a given v 0
what value of α 0 gives maximum height? What value gives maximum horizontal
range?
3. A monkey escapes from the zoo and climbs a tree. After failing to entice the
monkey down, the zookeeper fires a tranquilizer dart directly at the monkey (Fig.
3.13). The monkey lets go at the instant the dart leaves the gun. Show that the
dart will always hit the monkey, provided that the dart reaches the monkey before
he hits the ground and runs away.
4. For the rover of Examples 3.1 and 3.2, find the parallel and perpendicular
components of the acceleration at t=2.0 s .

5. Passengers on a carnival ride move at constant speed in a horizontal circle of

radius 5. 0 m, making a complete circle in 4.0 s. What is their acceleration?

62 | P a g e
 Figure 3.12 Our sketch for this problem #1 [1].

Figure 3.13 Our sketch for this problem #3 [1].

Figure 3.14 Our sketch for this problem #4: The parallel
and perpendicular components of the acceleration of the
rover at t=2.0 s . [1].

63 | P a g e
Exercise – MODULE 3
Name:__________________________________________________________Section:_____
II. Multiple Choice: Write the letter of your choice on the space provided. Then Explain
or show the solution that justifies your answer. Each item is worth 2 points; correct
answers but without solution will gain only 1 point.

_____1. A jet plane is flying at a constant altitude. At time it t 1=0 has components of
velocity v x =90 m/ s, v y =110 m/s . At time t 2=3.00 s the components are v x =−170 m/s ,
v y =40 m/s . For this time interval calculate the x -component of the average acceleration
a av− x.

a) - 8.67 m/s2 b) 8.67 m/s2 c) 7.68 m/s2 d) - 7.68 m/s2

_____2, Referring to problem number #1 calculate the y -component of the average

acceleration a av− y.

a) 2.33 m/s2 b) -2.33 m/s2 c) -2.35 m/s2 d) 2.35 m/s2

_____3. Referring to problem number #1 calculate the magnitude of the average

acceleration.
a) 8.24 m/s2 b) -8.24 m/s2 c) -8.98 m/s2 d) 8.98 m/s2

_____4. Referring to problem number #1 calculate the direction of the average

acceleration.

64 | P a g e
 a) 1950 b) 1930 c) 1920 d) 1910

_____5. A squirrel has x- and y-coordinates ( 1.1 m, 3.4 m ) at time t 1=0 and coordinates
( 5.3 m,−0.5 m ) at time t 2=3.0 s . For this time interval, find the x -components of the
average velocity v av− x.

a) 1.49 m/s b) 1.4 m/s c) 1.74 m/s d) 1.34 m/s

_____6. Referring to problem #5, for the given time interval, find the direction of the
average velocity.
a) 3170 b) 4170 c) 1170 d) 2170

_____7 Referring to problem #5, for the given time interval, find the magnitude of the
average velocity.
a) 1.9 m/s b) 1.8 m/s c) 1.7 m/s d) 1.6 m/s

_____8. Referring to problem #5, for the given time interval, find the y -components of
the average velocity v av− y .

a) 1.3 m/s b) -1.3 m/s c) 1.4 m/s d) -1.4 m/s

65 | P a g e
 _____9. A physics book slides off a horizontal table top with a speed of 1.10 m/s. It
strikes the floor in 0.350 s. Ignore air resistance. Find the height of the table top above
the floor.
a) 0.600 m b) 0.060 m c) 0.006 m d) 0.000600 m

_____10. Referring to problem #9, Find the horizontal distance from the edge of the
table to the point where the book strikes the floor;
a) 0.285 m b) 0.185 m c) 0.085 m d) 0.385 m

_____11. Referring to problem #9, find the horizontal components of the book’s velocity.
a) 3.10 m/s b) 2.10 m/s c) 1.10 m/s d) 4.10 m/s

_____12. Referring to problem #9, find the vertical components of the book’s velocity.
a) 3.43 m/s b) -3.43 m/s c) 3.45 m/s d) -3.45 m/s

_____13. Referring to problem #9, find the magnitude of its velocity, just before the book
reaches the floor.
a) 3.60 m/s b) -3.60 m/s c) -3.63 m/s d) 3.63 m/s

66 | P a g e
 _____14. Referring to problem #9, find the direction of its velocity, just before the book
reaches the floor.
a) 72.20 b) 71.20 c) 73.20 d) 75.20

Summary

Position, velocity, and acceleration vectors:

The position vector r⃗ of a point P in space is the vector from the origin to P . Its
components are the coordinates x , y , and z .
The average velocity vector ⃗v av during the time interval ∆ t is the displacement ∆ r⃗ (the
change in the position vector r⃗ ) divided by ∆ t . The instantaneous velocity vector ⃗v is the time
derivative of r⃗ , and its components are the time derivatives of x , y , and z . The instantaneous
speed is the magnitude of ⃗v . The velocity ⃗v of a particle is always tangent to the particle’s
path.
The average acceleration a av vector during the time interval ∆ t equals ∆ ⃗v (the change in
the velocity vector ⃗v ) divided by ∆ t . The instantaneous acceleration vector a⃗ is the time
derivative of ⃗v, and its components are the time derivatives of v x , v y and v z .
The component of acceleration a⃗ parallel to ⃗v the direction of the instantaneous velocity
affects the speed, while the component of perpendicular to affects the direction of motion.
Projectile motion:
In projectile motion with no air resistance, a x =0 and a y =−g. The coordinates and
velocity components are simple functions of time, and the shape of the path is always a
parabola. We usually choose the origin to be at the initial position of the projectile.
Motion in a Circle (Optional)
When a particle moves in a circular path of radius R with constant speed v (uniform
circular motion), its acceleration a⃗ is directed toward the center of the circle and perpendicular
to ⃗v.
If the speed is not constant in circular motion (nonuniform circular motion), there is still a
radial component of a⃗ given by Eq. (3-19) or (3-20), but there is also a component of a⃗

67 | P a g e
 parallel (tangential) to the path.

Post Test: This post-test can be conducted through online or face to face.

References
3. Kramer, Laird.Young, Hugh D. (2012) Study guide, Sears & Zemansky's University
physics, 13th edition, Young and Freedman /San Francisco, CA : Pearson.

MODULE 4

Newton’s Laws and Dynamics

Time frame: 5 hours

Objective: By studying this chapter you will learn:

 What the concept of force means in physics, and why forces are vectors.
 The significance of the net force on an object, and what happens when the net force is
zero.
 The relationship among the net force on an object, the object’s mass, and its
acceleration.
 How the forces that two bodies exert on each other are related.
 How to use Newton’s first law to solve problems involving the forces that act on a body
in equilibrium.
 How to use Newton’s second law to solve problems involving the forces that act on an
accelerating body.
 The nature of the different types of friction forces—static friction, kinetic friction, rolling
friction, and fluid resistance—and how to solve problems that involve these forces.

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  How to solve problems involving the forces that act on a body moving along a circular
path.
 The key properties of the four fundamental forces of nature.

Introduction:
Module 2 and 3 deal mainly on Kinematics that is, describing the motion of objects:
whether the object moves in one or two dimensions; whether it moves in a straight-line manner
or in a curve; whether it travels with constant velocity or with changing velocity. Equations of
motion have been derived based on definitions of displacement, velocity and acceleration; the
three physical quantities used to describe motion. In other words, the previous two modules
critically analyze how an object moves.
Module 4 answers the questions: What makes a body a resting body move or
accelerate? What causes the increase/decrease in the speed of a moving body? What causes
the change in the direction of a body’s motion? A critical answer to these questions are guided
by the Laws of Motion derived by Isaac Newton which use the physical quantity “force”. This
part of mechanics, which includes the study of both of motion and the forces that cause the
motion or change in body’s motion, is called dynamics.

Pre-test: This pre-test will be conducted through online or face to face oral recitation.

Learning Activities

A. Content/Discussion

FORCES
Force and Interactions
In everyday language, a force is a push or a pull. A better definition is that a force is
an interaction between two bodies or between a body and its environment Figure 4.1.

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 Figure 4.1 Some properties of forces.

When a force involves direct contact between two bodies, such as a push or pull that
you exert on an object with your hand, we call it a contact force. Figures 4.2a, 4.2b, and 4.2c
show three common types of contact forces.

Figure 4.2 Four common types of forces.

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 The normal force (Figure 4.2a) is exerted on an object by any surface with which it is in
contact. The adjective normal means that the force always acts perpendicular to the surface of
contact, no matter what the angle of that surface. By contrast, the friction force (Figure 4.2b)
exerted on an object by a surface acts parallel to the surface, in the direction that opposes
sliding. The pulling force exerted by a stretched rope or cord on an object to which it’s attached
is called a tension force (Figure 4.2c). When you tug on your dog’s leash, the force that pulls
on her collar is a tension force. In addition to contact forces, there are long-range forces that
act even when the bodies are separated by empty space. The force between two magnets is an
example of a long-range force, as is the force of gravity (Figure 4.2d); the earth pulls a dropped
object toward it even though there is no direct contact between the object and the earth. The
gravitational force that the earth exerts on your body is called your weight.

Superposition of Forces
When you throw a ball, there are at least two forces acting on it: the push of your hand
and the downward pull of gravity. Experiment shows that when two forces 1 and 2 act at the
same time at the same point on a body (Figure 4.3), the effect on the body’s motion is the same
as if a single force were acting equal to the vector sum of the original forces: ⃗
R =⃗
F 1 +⃗
F2 . More
generally, any number of forces applied at a point on a body have the same effect as a single
force equal to the vector sum of the forces. This important principle is called superposition of
forces.

Figure 4.4 Superposition of forces.

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 Figure 4.5 The force ⃗
F , which acts at an angle from the x -axis, may be replaced by its
rectangular component vectors ⃗
F x and ⃗
F y.

We will often need to find the vector sum (resultant) of all the forces acting on a body.
We call this the net force acting on the body. We will use the Greek letter Σ (capital sigma,
equivalent to the Roman S) as a shorthand notation for a sum. If the forces are labelled ⃗
F 1,⃗
F 2,
⃗
F 3 and so on, we abbreviate the sum as

⃗
R =⃗
F 1 +⃗
F2 + ⃗
F 3 +…=Σ ⃗
F (4-1)

R of two forces ⃗
Figure 4.6 Finding the components of the vector sum (resultant) ⃗ F1
and⃗
F 2.

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We read Σ ⃗
F as “the vector sum of the forces” or “the net force.” The component version of Eq.
(4-1) is the pair of component equations

R x =Σ F x
R y =Σ F y (4-3)

Once we have R x and R x we can find the magnitude and direction of the net force ⃗
R =Σ ⃗
F
acting on the body. The magnitude is

R=√ R x 2+ R y 2 (4-4)

Ry
and the angle between ⃗
R and the + x -axis can be found from the relationship tan θ= .The
Rx
components R x and R y may be positive, negative, or zero, and the angle may be in any of the
four quadrants. In three-dimensional problems, forces may also have z -components; then we
add the equation to Eq. (4.3) R z=Σ F z . The magnitude of the net force is then

R=√ R x 2+ R y 2+ R z2 (4-5)

NEWTON’S LAWS OF MOTION

Newton’s First Law

“A body acted on by no net force moves with constant velocity (which may be zero) and
zero acceleration”.
In other words when a body is either at rest or moving with constant velocity (in a
straight line with constant speed), we say that the body is in equilibrium. For a body to be in
equilibrium, it must be acted on by no forces, or by several forces such that their vector sum—
that is, the net force—is zero:

Σ⃗
F =0 (4-5)
(body in equilibrium)

Σ F x =0 , Σ F y =0 (4-6)

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 (body in equilibrium)

Newton’s Second Law

“If a net external force acts on a body, the body accelerates. The direction of
acceleration is the same as the direction of the net force. The mass of the body times the
acceleration of the body equals the net force vector” In symbols,

Σ⃗
F =m⃗a
(Newton’s second law of motion) (4-7)

There are at least four aspects of Newton’s second law that deserve special attention.
First, Eq. (4-7) is a vector equation. Usually we will use it in component form, with a separate
equation for each component of force and the corresponding component of acceleration:

Σ F x =m ax , Σ F y =m a y , Σ F z=ma z (4-8)
(Newton’s second law of motion)

Mass and Force

Experiments shows that the ratio of the magnitude of the net force to the magnitude of
the acceleration is constant, regardless of the magnitude of the net force. We call this ratio the
inertial mass, or simply the mass, of the body and denote it by m . That is,

|Σ ⃗
F| (4-9)
m=
a

Newton’s Third Law

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 “ If body A exerts a force on body B (an “action”), then body B exerts a force on body A
(a “reaction”). These two forces have the same magnitude but are opposite in direction. These
two forces act on different bodies.”

For example, in Fig. 4.7 ⃗

F A on B is the force applied by body A (first subscript) on body B (second
subscript), and ⃗
F B on A is the force applied by body B (first subscript) on body A (second
subscript). The mathematical statement of Newton’s third law is

⃗
F A on B =−⃗
F B on A (4-10)

(Newton’s third law of motion)

w=mg (4-11)
Free-body diagrams are essential to help identify the relevant forces. A free-body

diagram is a diagram showing the chosen body by itself, “free” of its surroundings, with vectors
drawn to show the magnitudes and directions of all the forces applied to the body by the various
other bodies that interact with it. We have already shown some free-body diagrams in Figures
4.8a, 4.8b, 4.8c. Be careful to include all the forces acting on the body, but be equally careful
not to include any forces that the body exerts on any other body. In particular, the two forces in
an action–reaction pair must never appear in the same free-body diagram because they never
act on the same body. Furthermore, forces that a body exerts on itself are never included, since
these can’t affect the body’s motion.

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 Figure 4.8 (a) The situation. (b), (c) Our free-body diagrams.

Frictional Forces
First, when a body rests or slides on a surface, we can think of the surface as exerting a
single contact force on the body, with force components perpendicular and parallel to the
surface (Figure 4.13). The perpendicular component vector is the normal force, denoted by n⃗ .
The component vector parallel to the surface (and perpendicular to n⃗ is the friction force,
denoted by ⃗f . If the surface is frictionless, then is zero but there is still a normal force. Again the
direction of friction force ⃗f is parallel with the surface of contact and opposite to the direction of
motion of the object.

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 Figure 4.9 When a block is pushed or pulled over a surface, the surface exerts a contact
force on it.

Kinetic friction force

The kind of friction that acts when a body slides over a surface is called a kinetic friction force
⃗f . Its magnitude is given by
k

f k =μk n (4-12)

Where μk is the coefficient of kinetic friction and n the normal force that acts on the objects.

Static friction force

Friction forces may also act when there is no relative motion. If you try to slide a box
across the floor, the box may not move at all because the floor exerts an equal and opposite
friction force on the box. This is called a static friction force ⃗f s. Its magnitude is given by

f s=μ s n (4-13)

Its values ranges from zero to a maximum, that is 0 ≤ f s ≥ μs n, where μs is called the coefficient
of static friction see Figure 4.13.

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 Figure 4.10 (a), (b), (c) When there is no relative motion, the magnitude of the static
friction f s force is less than or equal to μs n (d) When there is relative motion, the
magnitude of the kinetic friction force f k equals μk n.

Uniform Circular Motion

So far, the discussion on Newton’s Laws has focused on translational or straight-line
motion where a body accelerates due to it’s change in speed while moving in the same
direction. If a body moves with constant speed but constantly changes direction, the motion is
described as uniform circular motion. In this motion, the net Force acting on the body
constantly pulls it towards the center, thus naming it as centripetal force. Based on newton’s
2nd Law, if it follows that the body accelerates also towards the center with the magnitude,

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 2
v
a= , where v is the body’s speed and R is the radius of the circular path. The body in uniform
R
circular motion is illustrated in Figure 4.9.

Figure 4.11 A body that is moving in a uniform circular motion has a velocity with
constant magnitude but always changing in direction. The direction of the velocity ⃗v is
always tangent to the path of the body [1].

If the body has mass m, then then applying Newton’s Second law we obtain the net force to be

v2 (4-14)
Σ F=ma=m
R

Where R is the radius of the circular path.

SOLVING PROPBLEM WITH NEWTON’S LAWS

Using Newton’s Law: Particles in Equilibrium

Example 4.1: Two-dimensional equilibrium: SOLUTION:

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 In Figure 4.8, a car engine with weight w
hangs from a chain that is linked at ring O to
two other chains, one fastened to the ceiling
and
the other to the wall. Find expressions for the
tension in each of the three chains in terms
of w . The weights of the ring and chains are
negligible compared with the weight of the
engine.
Find expressions for the tension in each of EXECUTE: The forces acting on the engine
the three chains in terms of w . are along the y -axis only, so Newton’s first
law says:

Engine:
Σ F y =T 1 + (−w )=0
and
T 1=w , Eq.1
Ring: (FBD Figure 4.8c)
Σ F x =T 3 cos θ+ ( −T 2 ) =0, Eq.2
Σ F y =T 3 sin θ+ (−T 1 )=0 , Eq.3

Because T 1=w (from the engine equation),

we can rewrite the Eq.3 as
T1 w w
T 3= = = =1.2 w
sinθ sin θ sin 60 °

w cos 60 °
T 2=T 3 cos θ= cos θ=w =0.58 w
sin θ sin 60 °
Thus we obtained the expressions for the
tension T 1, T 2 and T 3 in each of the three
chains in terms of w .

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 Example 4.2: An inclined
plane: Figure 4.12 A cable holds a car at rest on a ramp.

A car of weight w rests on a

slanted ramp attached to a
trailer (Figure 4.12). Only a
cable running from the trailer to
the car prevents the car from
rolling off the ramp. (The car’s
brakes are off and its
transmission is in neutral.) Find
the tension in the cable and
the force that the ramp exerts
on the car’s tires.

Find the tension in the cable and

the force that the ramp exerts on Solution:
the car’s tires. EXECUTE: In Figure 4.12b we can see in the FBD the
forces acting on the Car on ramp

Σ F x =T +(−w sin α )=0 , Eq.1

Σ F y =n+(−w cos α )=0, Eq.2

Solving these equations for T and n , we find

T =w sin α
n=w cos α
Thus we obtained the expression for the tension in the
cable and the force that the ramp exerts on the car’s tires
which is the normal force n .

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Using Newton’s Second Law: Dynamics of Particles

Example 4.3: Straight-line motion with a Figure 4.13 (a) The situation. (b) Our free
constant force body diagram.

An iceboat is at rest on a frictionless

horizontal surface (Figure 4.13). A wind is
blowing along the direction of the runners so
that 4.0 s after the iceboat is released, it is
moving at 6.0 m/ s (about 22 km/h ,∨13 mi /h
). What constant horizontal force F W does
the wind exert on the iceboat? The combined
mass of iceboat and rider is 200 kg .

What constant horizontal force F W does the SOLUTION:

wind exert on the iceboat? The combined Execute: Using Newton’s Second Law we
mass of iceboat and rider is 200 kg . have

Σ F x =FW =m ax , Eq.1

Because here is no vertical acceleration, so

we expect that the normal force on the
iceboat is equal in magnitude to the iceboat’s
weight. That is,

Σ F y =n=w=mg=0, and
n=mg , Eq.2

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 The third equation is the constant-
acceleration relationship,
Eq. (2-21):
v x =v 0 x +a x t

v x −v 0 x
ax=
t

m m
6.0 −0.0
s s
a x= =1.5 m/s
4.0 s
From Eq. 1, F W =( 200 kg )( 1.5 m/s )

F W =300 kg ∙ m/s=300 N

Thus the constant horizontal force F W that the

wind exerted on the iceboat is 300 N .

Example 4.4: Tension in an elevator cable Figure 4.14 (a) The situation. (b) Our free
body diagram.
An elevator and its load have a combined
mass of 800 kg (Figure 4.14). The elevator is
initially moving downward at 10 m/s it slows
to a stop with constant acceleration in a
distance of 25.0 m. What is the tension T in
the supporting cable while the elevator is
being brought to rest?

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 What is the tension T in the supporting cable SOLUTION:
while the elevator is being brought to rest?
Execute: First let’s write out Newton’s second
law. The tension force acts upward and the
weight acts downward, so

Σ F y =T +(−w)=ma y, Eq.1
T =m a y +w
T =m a y +mg
T =m ( a y + g ) , Eq.2

To determine a y , we rewrite the constant

acceleration equation . Note that, the elevator
is moving downward with decreasing speed,
so its acceleration is upward; we chose the
positive y -axis to be upward.

2 2
v y −v 0 y =2 a y ( y− y 0 )
v y 2−v 0 y2
a y=
2( y− y 0 )
2 2
0.0 m/s −(−10.0 m/s )
a y=
2((−25)−0)
a y =+2.00 m/s 2
Now evaluating Eq.2 using the value of a y
we have
T =m ( a y + g )

T =800 kg ( + 2.00 m/ s +9.8 m/s )

2 2

T =9440 N .
Thus the tenstion force supporting cable
while the elevator is being brought to rest is

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 equal to 9440 N.

Example 4.5 Two bodies with the same acceleration

You push a 1.00−kg food tray through the cafeteria line with a constant 9.0−N force. The
tray pushes on a 0.50−kg carton of milk (Figure 4.15a). The tray and carton slide on a
horizontal surface so greasy that friction can be neglected. Find the acceleration of the tray
and carton and the horizontal force that the tray exerts on the carton.
Figure 4.15

Find the acceleration of the tray and SOLUTION

carton. Execute: Method 1: The x−¿ component
equations of Newton’s second law are
Note that F T on C =F C onT
Tray:
Σ F x =F+(−FC onT )=mT ax
and since
F T on C =F C onT
Thus,
Σ F x =F=F T onC +m T a x Eq.1

Milk Carton:

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 Σ F x =FT a x , Eq.2

Therefore from Eq.1 and Eq.2

F=m C a x +mT a x
F=a x ( mC + mT )

F 9.0 N 2
ax= = =6.0 m/s
( m C +mT ) 0.50 kg+1.00 kg

Find the horizontal force that the tray Substituting this value a x =6.0 m/s
2
into the
exerts on the carton. carton equation (Eq.2 ) gives

Σ F x =FT a x

F T on C =( 0.50 kg ) ( 6.0 m/ s2 ) =3.0 N

By using other method of finding the Method 2: The x−¿ component of Newton’s
acceleration of the tray and carton and the second law for the composite body of mass m is
horizontal force that the tray exerts on the Σ F x =F=ma x
carton.

F
a x=
m
Where m is the composite mass equal to 1.5 kg
9.0 N
ax= =6.0 m/ s2
1.5 Kg
And going back to Eq.2
Σ F x =FT a x

F T on C =( 0.50 kg ) ( 6.0 m/ s2 ) =3.0 N

Thus the answers are the same with both

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 methods.

Frictional Forces

Example 4.6 Friction in horizontal Figure 4.16 Our sketches for this problem [1].
motion

You want to move a500 N crate

across a level floor. To start the crate
moving, you have to pull with a
230 N horizontal force. Once the
crate “breaks loose” and starts to
move, you can keep it moving at
constant velocity with only 200 N .
What are the coefficients of static
and kinetic friction?
SOLUTION:
EXECUTE: Just before the crate starts to move (Fig.
4.16b), we have from Eq. (4-13),
f s=μ s n
now using newton’s first law and looking at the FBD at
Figure 4.14b we have,

Σ F x =T +(−f s )=0, Eq.1

and
T =f s=230 N ,
we also have
T =f s=μ s n, Eq.2.

Σ F y =n+(−w)=0 , Eq.3 and

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 n=w=500 N .

Thus we can obtain the coefficients of static friction

using Eq. 2 by substituting the value of n .
T 230 N
μs = = =0.46
n 500 N

After the crate starts to move (Figure 4.16c) we have

Σ F x =T +(−f k )=0,
T =f k =200 N ,
We also have
T =f k =μ k n , Eq.4.

Σ F y =n+(−w)=0 , and
n=w=500 N .

Thus we can obtain the coefficients of Kinetic friction

using Eq. 4 by substituting the value of n .

T 200 N
μk = = =0.40
n 500 N

As expected, the coefficient of kinetic friction is less

than the coefficient of static friction.

Uniform Circular Motion

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 Example 4.6 Rounding a flat curve
Figure 4.17 (a) The situation. (b) Our free
The sports car in Example 3.11 is rounding a body diagram [1].
flat, unbanked curve with radius R (Fig.
4.17a). If the coefficient of static friction μs
between tires and road is what is the
maximum speed v max at which the driver can
take the curve without sliding?

SOLUTION:
Execute:
The acceleration toward the center of the
2
v
circular path is a rad = . There is no vertical
R
acceleration. Thus we have

Σ F x =f =m arad , Eq.1
Σ F y =n+(−mg)=0, Eq.2

Since the car doesn’t slide toward or away

from the center of the circle, so the friction
force is static friction, with a maximum
magnitude f max =μ s n.

From equation Eq.1 we have

μs n=m arad, Eq. 3

From equation Eq.2 we have the value of n is

substituted to Eq.3 thus,

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 μs mg=ma rad

v2
μs g=arad =
R

2
v =μs gR
v max= √ μ s gR
As an example, if μs =0.96 and R=230 m we
have

v max= √( 0.96 ) ( 9.8 m/ s2 ) ( 230 m )=47 m/ s.

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B. Performance Task:
III. Problem solving: show your solutions completely and neatly.
1. A gymnast with mass m G=50.0 kg suspends herself from the lower end of a hanging
rope of negligible mass. The upper end of the rope is attached to the gymnasium
ceiling. (a) What is the gymnast’s weight? (b) What force (magnitude and direction)
does the rope exert on her? (c) What is the tension at the top of the rope?
2. Find the tension at each end of the rope in Problem #1 if the weight of the rope is
120 N.
3. Blocks of granite are to be hauled up a 15° slope out of a quarry, and dirt is to be
dumped into the quarry to fill up old holes. To simplify the process, you design a
system in which a granite block on a cart with steel wheels (weight w 1 including both
block and cart) is pulled uphill on steel rails by a dirt-filled bucket (weight w 1 including
both dirt and bucket) that descends vertically into the quarry (Figure 19). How must
the weights w 1 and w 2 be related in order for the system to move with constant
speed? Ignore friction in the pulley and wheels, and ignore the weight of the cable.
4. A 50.0-kg woman stands on a bathroom scale while riding in the elevator in Example
4.4. What is the reading on the scale?
5. A toboggan (Figure 4.20) loaded with students (total weight w ) slides down a snow-
covered slope. The hill slopes at a constant angle α , and the toboggan is so well
waxed that there is virtually no friction. What is its acceleration?
6. Figure 4.21 shows an air-track glider with mass m 1 moving on a level, frictionless air
track in the physics lab. The glider is connected to a lab weight with mass m2 by a
light, flexible, nonstretching string that passes over a stationary, frictionless pulley.
Find the acceleration of each body and the tension in the string.
7. In Example 4.6, suppose you move the crate by pulling upward on the rope at an
angle of 30 ° above the horizontall (Figure 2.22). How hard must you pull to keep it
moving with constant velocity? Assume that μk =0.40.

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 Figure 4.19 [1].
Figure 4.18 [1].

Figure 4.21 [1].

Figure 4.20 [1].

Figure 4.22 [1].

Exercise – MODULE 4
Name:________________________________________________________Section:_______
IV. Multiple Choice: Write the letter of your choice on the space provided. Then Explain or
show the solution that justifies your answer. Each item is worth 2 points; correct answers
but without solution will gain only 1 point.

_____1. Two 25.0-N weights are suspended at opposite ends of a rope that passes over
a light, frictionless pulley. The pulley is attached to a chain that goes to the ceiling. What
is the tension in the rope?
a) 26.0-N b) 27.0-N c) 25.0-N d) 28.0-N

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 _____2. Referring to Problem #1, what is the tension in the chain?
a) 20.0-N b) 30.0-N c) 50.0-N d) 40.0-N

_____3. A 125-kg (including all the contents) rocket has an engine that produces a
constant vertical force (the thrust) of 1720 N. Inside this rocket, a 15.5-N electrical power
supply rests on the floor. (a) Find the acceleration of the rocket.
a) 4.96 m/s2 b) 5.96 m/s2 c) 2.96 m/s2 d) 3.96 m/s2

_____4. Referring to Problem #3, when it has reached an altitude of 120 m, how hard
does the floor push on the power supply?
a) 22.7-N b) 21.7-N c) 23.7-N d) 24.7-N

_____5. In emergencies with major blood loss, the doctor will order the patient placed in
the Trendelenburg position, in which the foot of the bed is raised to get maximum blood
flow to the brain. If the coefficient of static friction between the typical patient and the
bedsheets is 1.20, what is the maximum angle at which the bed can be tilted with
respect to the floor before the patient begins to slide?
a) 200 b) 300 c) 400 d) 500
_____6. A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so
the box will slide out if it is set moving. The coefficients of kinetic and static friction
between the box and the bed of the truck are 0.355 and 0.650, respectively. Starting
from rest, what is the shortest time this truck could accelerate uniformly to 30.0 m/s
without causing the box to slide? Include a free-body diagram of the toolbox as part of
your solution.
a) 4.72 b) 4.71 s c) 4.73 s d) 4.74 s

_____7. A 25.0-kg box of textbooks rests on a loading ramp that makes an angle α with
the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static

93 | P a g e
 friction is 0.35. (a) As the angle α is increased, find the minimum angle at which the box
starts to slip.
a) 19.50 b) 19.30 c) 19.60 d) 19.90

_____8. Referring to Problem #7, at this angle, find the acceleration once the box has
begun to move.
a) 0.62 m/s2 b) 0.72 m/s2 c) 0.82 m/s2 d) 0.92 m/s2

_____9. Referring to Problem #7, at this angle, how fast will the box be moving after it
has slid 5.0 m along the loading ramp?
a) 3 m/s b) 6 m/s c) 5 m/s d) 4 m/s

_____10. A flat (unbanked) curve on a highway has a radius of 220.0 m. A car rounds
the curve at a speed of 25.0 m/s. What is the minimum coefficient of friction that will
prevent sliding?
a) 0.260 b) 0.290 c) 0.270 d) 0.280

_____11. Referring to Problem #10, suppose the highway is icy and the coefficient of
friction between the tires and pavement is only one-third what you found in part (a). what
should be the maximum speed of the car so it can round the curve safely?

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 a) 14.48 m/s b) 14.44 m/s c) 14.49 m/s d) 14.47 m/s

Summary

Force as a vector:
Force is a quantitative measure of the interaction between two bodies. It is a vector quantity.
When several forces act on a body, the effect on its motion is the same as when a single
force, equal to the vector sum (resultant) of the forces, acts on the body.
The net force on a body and Newton’s first law:
Newton’s first law states that when the vector sum of all forces acting on a body (the net
force) is zero, the body is in equilibrium and has zero acceleration. If the body is initially at
rest, it remains at rest; if it is initially in motion, it continues to move with constant velocity.
This law is valid only in inertial frames of reference.

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 Mass, acceleration, and Newton’s second law:
The inertial properties of a body are characterized by its mass. The acceleration of a body
under the action of a given set of forces is directly proportional to the vector sum of the forces
(the net force) and inversely proportional to the mass of the body. This relationship is
Newton’s second law. Like Newton’s first law, this law is valid only in inertial frames of
reference. The unit of force is defined in terms of the units of mass and acceleration. In SI
units, the unit of force is the newton (N)
Newton’s third law and action–reaction pairs:
Newton’s third law states that when two bodies interact, they exert forces on each other that
at each instant are equal in magnitude and opposite in direction. These forces are called
action and reaction forces. Each of these two forces acts on only one of the two bodies; they
never act on the same body.
Weight: The weight ⃗
w of a body is the gravitational force exerted on it by the earth. Weight is
a vector quantity. The magnitude of the weight of a body at any specific location is equal to
the product of its mass m and the magnitude of the acceleration due to gravity g at that
location. While the weight of a body depends on its location, the mass is independent of
location.
Friction and fluid resistance:
The contact force between two bodies can always be represented in terms of a normal
force n⃗ perpendicular to the surface of contact and a friction force ⃗f parallel to the surface.
When a body is sliding over the surface, the friction force is called kinetic friction. Its
magnitude f k is approximately equal to the normal force magnitude n multiplied by the
coefficient of kinetic friction μk . When a body is not moving relative to a surface, the friction
force is called static friction. The maximum possible static friction force is approximately equal
to the magnitude of the normal n force multiplied by the coefficient of static friction μs . The
actual static friction force may be anything from zero to this maximum value, depending on
the situation. Usually μs is greater than for a given pair of surfaces in contact.
Rolling friction is similar to kinetic friction, but the force of fluid resistance depends on the
speed of an object through a fluid.
Forces in circular motion:
In uniform circular motion, the acceleration vector is directed toward the center of the
circle. The motion is governed by Newton’s second law, ⃗
F =m ⃗a.

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Post Test: This post-test can be conducted through online or face to face.

References
4. Kramer, Laird.Young, Hugh D. (2012) Study guide, Sears & Zemansky's University
physics, 13th edition, Young and Freedman /San Francisco, CA : Pearson.

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