Chapter I

Measurement

Target Outcomes

At the end of the lesson, you are expected to:

1. solve measurement problems involving conversion of units, expression of measurements notation
2. transform given value into scientific notation
3. determine significant figures in a given value
4. perform operations regarding scientific notations

Abstraction

Measurement

Measurement is a quantitative description of a fundamental property such as length, mass, time, force, weight, volume,
temperature luminous intensity, etc. It is a process of associating numbers to compare one quantity with another quantity. In everyday
life, we encounter measurement in our various activities such as cooking, doing exercise, and even washing our clothes. In Physics, it
plays a very essential role in understanding phenomena especially when performing experiments.

Systems of Measurement
There are two systems that carry different standardized units: the English system (also called British Engineering System)
and the Metric System.
English System was first used in England and it is also known as FPS system (foot, pound, second). This is a product of
creative measurement using objects and human body parts as measuring devices. For example, distances were measured with the
human foot and capacities were measured with household items such as cups, pails (formerly called gallons) and baskets. It was
lawfully accepted for use in the United States in 1866, the US has not adopted the metric system as its "official" system of
measurement.

Metric System was originated from France in 1791 by the French Academy of Science. It was based on a new unit of
length metre derived from the Greek work metron which means measure. International System of Measurement (SI) is the
modernized version of it. It was fully adopted in the Philippines on January 1, 1983 under the Batas Pambansa No. 8 due to it
convenience and accuracy to use because it uses decimal or base-10 system.

Categories of Physical Quantity

1. Fundamental Quantities are basic measurable and independent physical quantities. The 7 fundamental quantities with their SI
base units are:
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2. Derived Quantities are combination of fundamental quantities. These cannot be measure directly but can be calculated in Physical
Science. Example of these are:

Conversion of Units
Example 1. What is the mass in kg of a person weighing 170 lbs?
Steps in converting one unit to another unit

1. Determine the conversion factors between the starting quantity (the one that is given in the problem) and relate to other
units of quantity you are working with. Refer to the conversion table of units.
170 lbs (given unit) convert to kg (desired unit)
1 kg= 2.2046 lbs
2. Transform the conversion factor into factor form. The numerator will be the desired unit while the denominator will be
the given unit.
1kg
2.2046 lbs
3. Multiply it out and cancel for any units that are both top and bottom.
170 lbs x 1kg = 77.11 kg
2.2046 lbs

Example 2. Covert 3 km/h to m/s

1. From km/h to m/h (1kilometer=1,000 meters)

3km x 1000 = 3000m

h 1km h

2. From m/h to m/s (1hour=3,600 seconds)

3000m x 1h = 3000m = 0.83 m/s

1h 3600s 3600s

Example 3. Covert 25 ⁰C to ⁰F

9
° F= C+32
5
9 0
° F= (25 C)+32
5
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Significant Figures are numerical values in a measurement that are known with certainty. They include all the digits precisely known
plus an estimated last digit.

Rules for Significant Figures

1. All non-zero numbers are significant. The number 33.2 has THREE significant figures because all of the digits present are
non-zero
2. Zeros between two non-zero digits ARE significant. 2051 has FOUR significant figures. The zero is between a 2 and a 5.
3. Leading zeros are NOT significant. The number 0.54 has only TWO significant figures. 0.0032 also has TWO significant
figures. All of the zeros are leading.
4. Trailing zeros to the right of the decimal ARE significant. There are FOUR significant figures in 92.00
5. Trailing zeros in a whole number with the decimal shown ARE significant."540." indicates that the trailing zero IS
significant; there are THREE significant figures in this value.
6. Trailing zeros in a whole number with no decimal shown are NOT significant. Writing just "540" indicates that the zero is
NOT significant, and there are only TWO significant figures in this value.
7. An over line, sometimes also called an overbar, may be placed over the last significant figure; any trailing zeros following
this are insignificant. For example, 1300 has three significant figures

Scientific Notation is the way that scientists easily handle very large numbers or very small numbers.

Exponent
Coefficient
1.23 x 10 4

Base
Rules in Writing Scientific Notation

1. The coefficient must be greater than 1 and less than 10 and contain all the significant digits in the number. 
 12.3x104  is not in scientific notation, since the coefficient is greater than 10 Neither is 
 0.123x107 since the coefficient is less than 1
2. Identify the first significant digit and place the decimal number to its right.
 347,000. will be 3.47 since it has 3 significant figures and the first one is 3
3. The base is always 10.
1.23 Base
4. The exponent is the number of places the decimal was moved to obtain the coefficient.
 If the decimal point is moved to the left, exponent is positive.
347,000 3.47 X 105
 If the decimal point is moved to the right, exponent is negative.
0.000882 8.82 X 10-4

Calculating with Scientific Notation

Rules for Multiplication:

a.) 3.45 x 107 x 6.25 x105

1. Rewrite the problem as: (3.45x6.25) (107+5 )

2. Multiply the coefficients and add the exponents: 21.5625 x 1012

3. Change to correct scientific notation and round off to correct significant figures. 2.16 x 1012

b.) 2.33 x10-6 x 8.19 x103

1. Rewrite the problem as: (2.33x8.19) (10-6 +3 )

2. Multiply the coefficients and add the exponents (subtract if unlike signs then follow the sign of the bigger value) 19.0827x 10-3

3. Change to correct scientific notation and round off to correct significant figures. 1.91 x 10-3

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Rules for Division

a.)3.5 x108 / 6.6 x104

1. Rewrite the problem as: (3.45/6.25) (108-4 )
2. Divide the coefficients and subtract the exponents: 0.530303 x 104
3. Change to correct scientific notation and round off to correct significant figures 5.30 x 10(4-1) =5.30 x 103

b.) 8 x104 / 2 x10-2

1. Rewrite the problem as: (8/2) (104-(-2))

2. Divide the coefficients and subtract the exponents: 4 x 106
3. Change to correct scientific notation and round off to correct significant figures. 4.0 x 106

Rules for Addition

a.) 3.76 x104 + 5.5 x102

1. Rewrite the problem as: 3.76 x104 + 5.5 x102
2. Move the decimal of one of the addends to make the exponents the same: 0.055 x 104
3. Add the coefficients and leave the base and exponents the same 3.76 +0.055 = 3. 815x104

Rules for Subtraction

b.) 9.7 x104 - 4.8 x105

1. Rewrite the problem as: 9.7 x104 + 4.8 x105
2. Move the decimal of one of the minuend to make the exponents the same:0.97 x 105
3. Add the coefficients and leave the base and exponents the same
4.8-0.97 = 3. 83 x105

Utilization of Learning

I. Convert each quantity to the given units. Present all your II. Determine the number of significant figures in each item.
solutions. Use short bond paper for your solution. Write your answer after each item.

1. 280 in= _____________ ft 1. 246.32 mL

2. 26 T = _____________ lb 2. 107.854 km

3. 11 yd= _____________ ft 3. 100.3 g

4. 5 mi2 = _____________ acres 4. 0.678 mi

5. 56 gal = _____________ fl oz 5. 1.008 ft

6. 78 L= _____________ qt 6. 0.00340 lbs

7. 38 mi/h= ___________ km/h 7. 14.600 m

8. 88.5 ⁰C= ____________ ⁰F 8. 0.0001 inch

9. 78.4 ⁰F= ____________ ⁰C 9. 350.670 tons

10. 2020 m/s = ____________ km/h 10. 1,0000 yd

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III. Direction: Perform the indicated operation. Write your 5. 5.3× 105 + 6.452 × 102
answer after each item. Use short bond paper for your
solution. 6. 2.71× 107 – 8.77 × 105

1. (1.08 × 10-3)(9.3 × 10-3) 7. (8.02 × 10-5)(3.2 × 10-3) (5.8 × 104)

2. 7.1 × 108 / 8.2 × 106 8. (5.342 × 10-6 x 4.5 × 10-4) / 6.6 × 106

3. (2 × 10-5)(8.1 × 10-2) 9. 9.347× 1010 + 7.837 × 108

4. 4 × 105/ 3.63 × 10-3 10. 3.35× 107 – 7.67 × 105

IV .Make a paper airplane and test it out 3 times. Measure the distance after each throw using inches and centimeters. Record
the data then answer the questions below. Use short bond paper for your solutions.

Trial Distance
Inches Centimeters
1
2
3

1. What is your farthest throw? Convert it into feet and meter.

_________________ inches/cm = ____________________ m

2. Add all your throws together. How far has your paper airplane travelled in yard?
__________________ inches = _______________________ m
__________________ cm = _______________________

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 Chapter II
Vectors

Target Outcomes

At the end of the lesson, you are expected to

1. analyze the difference between Vector and Scalar quantities

2. perform addition of vectors
3. rewrite a vector in component form
4. perform operations regarding vectors using trigonometric functions
5. calculate directions and magnitudes of vectors

Abstraction

 Scalar Quantities are quantities are measured with numbers and units. Distance vs Displacement
Examples of scalars are distance, length, speed, time, mass,
temperature and energy. Vector Quantities are quantities which
require both magnitude and direction for their complete description.
Displacement, velocity, acceleration and force are examples of vectors.

Vector Representation

A vector quantity is represented by an arrow called vector ( )

Parts of a Vector
1. Arrowhead- indicates direction

2. Length of the arrow- indicates magnitude

3. Tail- represents point of origin

Vectors can be represented in a boldface type or by placing an arrow over the symbol. Velocity and displacement have arrows
above their symbols because they are vector quantities. Force is an example of vector also.

v=s/t F=m a

Reference Frame pertains to a coordinate system or sets of axes within which is used to measure the position, orientation and
other properties of an object in it. For example, a person standing on the side of the road watching a car drive past him from left
to right. In his frame of reference, the spot where he is standing is the point of origin, the road as the x-axis and the direction in
front of him is the y-axis. Direction guide and Cartesian coordinate plane are the two common reference frames used by vector
quantities.

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 How to describe Vector’s direction?

Operations with Vectors

Vectors that add together are called component vectors while the sum of component vectors is called resultant R. The process of
finding the effective value of a component vector in a given direction is called the resolution of vectors.

General Rules in finding the Resultant Vector

1. Vectors in the same direction are added while vectors in opposite
direction are subtracted.

2. When vectors acting at right angle with each other, use

Pythagorean Theorem.
For example:
A student walks 10m East then turns and walks 15m
North to reach his friend’s house from her place. Where is
the student with respect to his house?
N
Solution:
a 2 + b2 = h2
(15 m)2 + (10 m)2 = h2
R=? 15 m 225 m2 + 100 m2 = h2
√325 m2 =√ h2
18.027 m= h
E
10 m

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 R=18.027 m, 53.31⁰ N
For the angle of location, we will use tangent=opposite side/adjacent side (SOHCAHTOA)

N
tanꝊ= 15 m
10 m
tanꝊ= 1.5 m
Ꝋ=tan-1 1.5m
Ꝋ=53.31⁰
R=18.027 m, 53.31⁰ N 53.31⁰
E

3. When vectors acting at an angle, use Cosine law.

For example:
An insect crawls 18ft westward direction then runs 20ft, 37⁰ Northwest. If the insect is
approximately 34 ft from the point of origin, where is it located?

Given:
a=34 ft, b=20 ft, c= 18 ft Solution:
b2 = a2+c2-2accosB
N (20ft)2=(34ft)2+(18ft)2-2(34ft) (18ft)cosB
400ft2=1156 ft2+324 ft2-2(612 ft2) cosB
400ft2=1480 ft2-2(612 ft2) cosB
400ft2=1480 ft2-1224 ft2 cosB
C 400ft2-1480 ft2=-1224 ft2 cosB
-1080 ft2 = -1224 ft2 cosB
-1224 ft2 -1224 ft2
a=34 ft 0.8824 =cosB
b=20 ft cos0.8824 =B
28.07⁰=B

W B=28.07⁰ E
A c=18 ft

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 Chapter III
KINEMATICS: Motion along a Straight Line

Target Outcomes

At the end of the lesson, you are expected to:

1. classify physical situations into speed and velocity
2. transform a given problem into mathematical expression to describe motion
3. solve for unknown quantities in physical situations involving speed, velocity one-dimensional accelerated motion and free fall

Abstraction

Kinematics is a branch of Physics that describes the quantitative description of motion such as the rate at which the particles are
moving (velocity) and the rate at which their velocity is changing (acceleration).

Motion is the change of position with respect to a certain reference point. It is an object’s displacement in accordance to objects that
are considered to be stationary.

Examples of Motion:

What is a Reference Frame?

It is a point used to define the location of another point. In the illustration below, the boy’s displacement is +8 East in accordance to
Alex’s house. If the city hall is the reference frame, his displacement will be +12 East.

Motion in a
straight line

Rectilinear or Translatory motion is the motion of a body along a straight line. It is the most basic of all motion. A ball thrown
straight upward and falling back straight down is an example of it.
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 Speed vs Velocity

Speed is a scalar quantity which specifies magnitude of the rate of motion without reference to the direction of motion. It is the
distance traveled per unit of time. Velocity is the total displacement covered in a given time interval. It is the speed in a given
direction.

(d)distance Velocity ( v ) =
(d)displacement
Speed ( v ) = (t)time
(t)time

Sample problem

A car has traveled a distance of 30 meters in 10 seconds from point A to point B. What is the average speed of the car? If the
displacement if the car is 20 meters North, what’s the car’s average velocity?

Given: Find: Solution:

d 30 m
distance (d) =30 m v=? if d = 20 m North Speed: v= = =3 m/s
t 10 s
time (t) = 10 seconds v=? Velocity:
d 20 m North 2 m
v= = = North
t 10 s s

Given: Find:
A car travelled East 20 m/s for 35 s, North at 32 m/s for 40 s and West 45 m/s for 50 s. v1=20 m/s dtotal
v2=32 m/s d
a. How far did it travel? Given: Given: v3=45 m/s Find: Find: v
b. What was its displacement during the entire trip? v1=20 m/s v1=20 m/s t1 = 35 s dtotal dtotal vave
c. Find its average speed v2=32 m/s v2=32 m/s t2 = 40 s d d
v3=45 m/s v3=45 m/s t3 = 50 s v v
t1 = 35 s t1 = 35 s vave vave
N
t2 = 40 s t2 = 40 s 2250 m
t3 = 50 s t3 = 50 s
Solution:
d 1280 m
1280 m
d1=v1t1=(20 m/s)(35 s)= 700 m East
d2=v2t2=(32 m/s)(40 s)= 1280 m North
d3=v3t3=(45 m/s)(50 s)= 2250 m West

ttotal= t1+t2+ t3
ttotal=35 s + 40 s +50 s
ttotal=125 s E
1550 m 700 m

a. dtotal= d1 +d2 +d3

dtotal=700 m+1280 m+2250 m
dtotal=4230 m

b. Pythagorean Theorem 1280𝑚 d=?

𝑡𝑎𝑛𝜃 = 1280 m
d= √( 1280 ) + ( 1550 )
1550𝑚
2 2 −1
𝜃 = 𝑡𝑎𝑛 0.83
𝜃 = 39.6⁰
d=2010 m

𝑑 total 𝑑 d d
a. 𝑣 =
𝑡 total
a.c. 𝑣 = 𝑡 total d. 𝑣𝑎𝑣𝑒 = 𝑡 d. 𝑣𝑎𝑣𝑒 = 𝑡
total total total
4230 𝑚 4230𝑚 2010 𝑚 2010 𝑚 d=2010 m, 𝟑𝟗. 𝟔𝟎 𝑵𝒐𝒓𝒕𝒉 𝒐𝒇 𝑾𝒆𝒔𝒕
𝑣= 𝑣= 𝑣𝑎𝑣𝑒 = 𝑣𝑎𝑣𝑒 =
125s 125s 125𝑠 125𝑠
10 |𝒎/𝒔
𝒗 = 𝟑𝟑 G e n e r a𝒗 l= P
𝟑𝟑h 𝒎/𝒔
y s i c s 𝒗1 =b𝟏𝟔.
y 𝟏L𝒎/𝒔
. A . R𝒗
o t=o𝟏𝟔.
n i 𝟏 𝒎/𝒔
Acceleration measures the rate of change in velocity. When a body’s velocity increases, it is said that the body accelerates. On the
other hand, when a body’s velocity decreases, it is said that the body decelerates. When a body’s acceleration is constant, it is said to
be in uniformly accelerated motion. The following formula can be used for obtaining acceleration:

∆v v −v
a= ∨a= f i
t t
Where:
m
a=acceleration( 2
)
s
t=time( s)
∆ v=change ∈velocity (m/s)
v f =final velocity( m/s)
vi =initial velocity (m/s )

Sample Problem:

1. A car accelerates uniformly from 20 m/s to 50 m/s in 6 s. What is its acceleration?

Given: Solution:
50m/ s−20 m/s
v f =50 m/ s a=
6s

30 m/ s
vi =20 m/s a=
6s
2
t=6 s a=5 m/ s

2. An objects decelerates uniformly from 60 m/s to a stop in 10 s. What is its acceleration?

Given: Solution:
0 m/s−60 m/s
v f =0 m/s a=
10 s

−60 m/s
vi =60 m/s a=
10 s
t=10 s a=−6 m/s
2

For any problems on uniformly accelerated motion, the following formulas are applicable:
𝒂𝒕𝟐
𝒅 = 𝒗𝒊 𝒕 +
𝟐 m
Where: a=acceleration( )
𝒗𝒇 = 𝒗𝒊 + 𝒂𝒕
s2
𝒗𝒇𝟐 = 𝒗𝒊𝟐 + 𝟐𝒂𝒅
𝒗𝒇 + 𝒗𝒊
𝒗=
𝟐
𝒅 = 𝒗𝒕

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 t=time elapsed (s)
v=average velocity (m/s)
v f =final velocity( m/s)
vi =initial velocity (m/s )
d=displacement (m)

Sample Problem
Car moving in a straight line at 12 m/s accelerates uniformly at 2.50 m/s2.
a. How fast will it go at the end of 7 s?
Given: Find: vf when t=7 s Solution:
a = 2.50 m/s2 v f =v i +at
2
vi = 12 m/s v f =12 m/s +2.50 m/s ¿)
m m
v f =12 +17.5
s s
v f =29.5m/ s
b. How far will it have gone in 9 s?
Find: t when d=100 m Solution:
2
at
d=v i t+
2
m 2
(2.50 2
)(9 s )
m s
d=(12 )( 9 s)+
s 2
d=108m+101.25 m
d=209.25m

c. In how many seconds will it be 100 m from the starting point?

Find: t when d=100 m Solution:

v f =v i + 2 ad v f −vi
2 2
a=
t
v f =¿
2
m
2 2 25.4 −12 m/s
m m s
v f =144 +500 2 2.50 m/s 2=
t
2
2
s s
m
2 13.4 m/s
√ v f =√ 644 2.50 m/s 2=
2

s2 t
13.4 m/s
Free Fall is any motion of a body where gravity is the only acceleration acting upon it.t=It is up or down motion.
2 On or near the
2.50
Earth’s surface, the acceleration due to gravity of an object is approximately 9.8 m/s2 or 32 ft/ m/s
s2. The motion of a free falling body is
not affected by mass, size and shape. An object has an increasing acceleration as it falls to the ground. Hence, two objects of different
masses dropped from the same height above the Earth’s surface will fall to the ground at the same time in the absence of air
resistance.

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 Examples of Free Fall

The following formulas can be used in solving problems on free falling bodies:
gt2
d=v i t+
2
v f =v i +¿
v f =v i + 2 gd
2 2

Where:
m '
g=acceleration due ¿ gravity (−9.8 2
if near the Eart h ssurface)
s
t=time elapsed (s)
v f =final velocity( m/s)
vi =initial velocity (m/s )
d=displacement=shortest vertical distance ¿ the initial ¿ final position( m)
The following sign convention is followed when g used is negative:
1. vi is + when the initial direction is vertically upward
2. vi is - when the initial direction is vertically downward
3. v f is + when the direction of the object at this final velocity is vertically upward
4. v f is - when the direction of the object at this final velocity is vertically downward
5. d is + when the final location being considered is above the initial location
6. d is - when the final location being considered is below the initial location
Note that the sign convention is reversed when g used is positive. In the problems that we are going to consider, we will assume that
the activity is done near the Earth’s surface unless otherwise stated.

Sample Problem:
A stone is thrown vertically upward from the ground with a velocity of 20 m/s
a. How high will it rise?
b. In how many seconds will it reach its maximum height?
c. With what velocity will it strike the ground?
d. In how many seconds will it strike the ground?
Solution:
m
a. vi =20
s Note: At maximum height the

v f =v i + 2 gd
2 2
vf will be 0 since it can’t rise
any farther. The final location of
02 =¿ the object is above the initial
m2 m2 position so d will become +
g=-9.8 m/s2 d=? 0=400 2 −19.6 2 (d )
s s
2 2
m m
19.6 2 (d )=400 2
s s
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 2
m
400
2
s
d= 2
m
19.6 2
s
d=20.44 m
a.
b. v f =0
z v f =v i +¿
0=20 m/s+ ¿) t
m
9.8 2 (t )=20 m/s
g=-9.8 m/s2 s

20 m/s
t=
m
9.8 2
s

v f =v i + 2 gd
2 2
c.
Note: The d is 0 since it
v f =¿
2 can’t go any further when it
reached the ground. We
2
m choose – since the direction
g=-9.8 m/s2 √ v f =± √ 400
2
2 at its final position is
s downward. vi and vf have
d=0, vf=?
v f =−20 m/s the same magnitude because
they are at the same level.
m
v f =20 downward
s

d. g t2
d=v i t+
2
m 2
(−9.8 2
)t
s Note: The time to reach the
0=20 m/s (t)+ ground is twice that to
g=-9.8 m/s2
2 reach the maximum height.
d=0, t=? m 2
(−9.8 2 )t
s
20 m/s (t)=
2
m 2
(−9.8 2 )t
s
20 m/s(t ) 2
=
t t

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 end

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