‫الجامعة السعودية االلكترونية‬

‫الجامعة السعودية االلكترونية‬

‫‪26/12/2021‬‬
College of Computing and Informatics
Data Science Pre-Master Program
SCI 101
General Physics I
SCI 101
General Physics
Week 2
Motion in one dimension
Contents

2.1 Position, velocity and speed of a particle

2.2 Instantaneous Velocity and Speed

2.7 Particles under constant acceleration

2.8 Freely Falling objects

Weekly Learning Outcomes
• Recall basic and derived physical quantities

• Differentiate distance and displacement, speed and velocity

• Understand motion with constant acceleration

• State freely falling motion and solve related problems

Required Reading

Chapter 2 Motion in One Dimension in Physics for scientists and

engineers with modern physics

Recommended Reading
Chapter 2 Motion Along a Straight Line in Principles of
Physics, 10th Edition. John Wiley & Sons
 Recommended Video
Motion along one dimension https://www.youtube.com/watch?
v=EhsEIRAImM8&list=PLkuVD5NmHtHfEUA4hNycY-YqPFEK5dugn&index=4

Motion along straight line https://www.youtube.com/watch?

v=EhsEIRAImM8&list=PLkuVD5NmHtHfEUA4hNycY-YqPFEK5dugn&index=3

This Presentation is mainly dependent on the textbook: Data Structures and Algorithm Analysis in Java by Mark Allen Weiss
• Position, velocity and speed of a particle
Linear Speed
• Speed is a “scalar quantity”
• does not include direction.
• I am running at 10 mph, but I do not tell you
where – this is speed.
• Speed is my distance covered divided by the time
it takes me
Linear Speed
Average Speed
• It would take us a lot longer than 15 minutes to drive
downtown because of the traffic where we have to
slow, stop, start, accelerate, slow, stop, etc., etc,.
Linear Speed
Average Speed
• Therefore in planning a trip we think about average
speed
• Average Velocity
• Total distance (d) covered/ time (t)
v = d2 – d1
t
 Average Velocity
Speed is how far an object travels in a given time interval:

Velocity includes directional information:

 Average Velocity
Example 2-1: Runner’s average velocity.
The position of a runner as a function of time is plotted as
moving along the x axis of a coordinate system. During a 3.00-s
time interval, the runner’s position changes from x1 = 50.0 m to
x2 = 30.5 m, as shown. What was the runner’s average velocity?

Figure Caption: Example 2–1.

A person runs from x1 = 50.0 m to x2 = 30.5 m. The displacement
is –19.5 m.
Answer: Divide the displacement by the elapsed time; average
velocity is -6.50 m/s
 Instantaneous Velocity
The instantaneous velocity is the average velocity in the
limit as the time interval becomes infinitesimally short.

Ideally, a speedometer would measure

instantaneous velocity; in fact, it measures
average velocity, but over a very short time
interval.
Car speedometer showing mi/h in white, and km/h in
orange.
 Instantaneous Velocity

The instantaneous speed always equals

the magnitude of the instantaneous
velocity; it only equals the average
velocity if the velocity is constant.

Velocity of a car as a function of time: (a) at constant velocity; (b) with varying velocity
 Quick check

• Going in a straight line at the same speed is called

Constant Velocity.
• But if we change our speed (brake at a light) or our
direction (turn a corner) then we have a Changing
Velocity
• Changing our speed or direction is Acceleration
 Acceleration
Acceleration is the rate of change of velocity.

Example : Average acceleration.

A car accelerates along a

straight road from rest to 90
km/h in 5.0 s. What is the
magnitude of its average
acceleration?
Example 2–4.The car is shown at the start with v1 = 0 at t1 = 0. The
car is shown three more times, at t = 1.0 s, t = 2.0 s, and at the end of
our time interval, t2 = 5.0 s. We assume the acceleration is constant
and equals 5.0 m/s2. The green arrows represent the velocity vectors;
the length of each arrow represents the magnitude of the velocity at
that moment. The acceleration vector is the orange arrow. Distances
are not to scale.
• Solution: The average acceleration is the change in speed divided
by the time, 5.0 m/s2.
 Acceleration
Conceptual Example : Velocity and acceleration.
(a) If the velocity of an object is zero, does it mean that the
acceleration is zero?
(b) If the acceleration is zero, does it mean that the velocity is zero?
Think of some examples.
Linear Acceleration
• Acceleration is how quickly velocity changes
a = v2 – v1
t
• When we accelerate in a car from stop to 60km/h
in 5 seconds.
a = (60 km/h – 0 km/h)/5s
= 12 km/h/s
 Acceleration
Example : Car slowing down.
An automobile is moving to the right along a straight highway,
which we choose to be the positive x axis. Then the driver puts on
the brakes. If the initial velocity (when the driver hits the brakes) is
v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what
was the car’s average acceleration?

showing the position of the car at times t1 and t2, as well

as the car’s velocity represented by the green arrows. The
acceleration vector (orange) points to the left as the car
slows down while moving to the right.

Solution: The average acceleration is the change in speed

divided by the time; it is negative because it is in the
negative x direction, and the car is slowing down: a = -2.0
m/s2
 Acceleration
There is a difference between negative acceleration and
deceleration:
Negative acceleration is acceleration in the negative direction as
defined by the coordinate system.
Deceleration occurs when the acceleration is opposite in direction to
the velocity.

The car now moving to the left and decelerating. The acceleration is +2.0 m/s.
 Acceleration
The instantaneous acceleration is the average acceleration in the
limit as the time interval becomes infinitesimally short.

A graph of velocity v vs. time t. The average acceleration

over a time interval Δt = t2 – t1 is the slope of the straight
line P1P2: aav = Δv/ Δt. The instantaneous acceleration at
time is t1 the slope of the v vs. t curve at that instant.
 Acceleration
Example : Acceleration given x(t).
A particle is moving in a straight line so that its position is given by the relation x = (2.10
m/s2)t2 + (2.80 m). Calculate (a) its average acceleration during the time interval from t1 =
3.00 s to t2 = 5.00 s, and (b) its instantaneous acceleration as a function of time.

Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the motion x = At2 + B. Note that increases linearly with and that the
acceleration a is constant. Also, v is the slope of the x vs. t curve, whereas a is the slope of the v vs. t curve.
Solution: The velocity at time t is the derivative of x; v = (4.20 m/s2)t.
a. Solve for v at the two times; a = 4.20 m/s2.
b. Take the derivative of v: a = 4.20 m/s2.
 Motion at Constant Acceleration

The average velocity of an object during a time interval t is

The acceleration, assumed constant, is

 Motion at Constant Acceleration

In addition, as the velocity is increasing at a constant rate, we know

that

Combining these last three equations, we find:

 Motion at Constant Acceleration

We can also combine these equations so as to eliminate t:

We now have all the equations we need to solve constant-

acceleration problems.
 Freely Falling Objects
Near the surface of the Earth, all objects experience approximately
the same acceleration due to gravity.

This is one of the most common

examples of motion with constant
acceleration.

Multiflash photograph of a falling apple, at equal time intervals. The

apple falls farther during each successive interval, which means it is
accelerating.
 Freely Falling Objects

The acceleration due to gravity at

the Earth’s surface is approximately
9.80 m/s2. At a given location on the
Earth and in the absence of air
resistance, all objects fall with the
same constant acceleration.

A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum.
Freely Falling Objects

Example : Falling from a tower.

Suppose that a ball is dropped (v0 = 0)
from a tower 70.0 m high. How far will it
have fallen after a time t1 = 1.00 s, t2 =
2.00 s, and t3 = 3.00 s? Ignore air
resistance.

Example (a) An object dropped from a tower falls with progressively

greater speed and covers greater distance with each successive
second. (See also Fig. 2–26.) (b) Graph of y vs. t.

Solution: We are given the acceleration, the initial speed, and the time;
we need to find the distance. Substituting gives t1 = 4.90 m, t2 = 19.6
m, and t3 = 44.1 m.
 Freely Falling Objects

Example : Ball thrown upward, II.

Let us consider again a ball thrown upward with
initial velocity of 15 m/s, and make more
calculations. Calculate (a) how much time it takes
for the ball to reach the maximum height, and (b)
Find the maximum height of the stone (c)
Determine the velocity of the ball when it returns
to the thrower’s hand (point C).
a) When the ball reaches its maximum
height, its velocity becomes zero b) The maximum height is obtained from the following formula

v0  15m / s 1 2
y  y0  v0t  at
v0 2
v  v0  at 1
y  0  15(1.53)  (9.8)(1.53) 2
v  v0  gt 2
y  11.48m
0  15  (9.8)t
15
t  1.53
9.8
c) The maximum height is obtained from the following formula
c) Notice that the initial and final position of the ball is zero so we have

v  v  2a ( y f  y i )
2 2
0

v 2  (15) 2  2(9.8)(0  0)
v 2  225
v   225  15m / s
 Freely Falling Objects
Example : Ball thrown upward, III; the quadratic formula.
For a ball thrown upward at an initial speed of 15.0 m/s, calculate at
what time t the ball passes a point 8.00 m above the person’s hand.

Graphs of (a) y vs. t, (b) v vs. t for a ball thrown upward,

Solution: We are given the initial and final position, the initial speed, and the acceleration, and want to find the
time. This is a quadratic equation; there are two solutions: t = 0.69 s and t = 2.37 s. The first is the ball going up
and the second is the ball coming back down.
 1 2
y  y0  v0t  at
2
1 2
8  0  15t  gt
2
9.8 2
8  15t  t
2
4.9t  15t  8  0
2

15  68.2
t
9.8
t1  0.695s
t2  2.37 s
 Freely Falling Objects
Example : Ball thrown upward at edge of cliff.

Suppose that a ball is thrown upward at a speed of

15.0 m/s by a person standing on the edge of a cliff,
so that the ball can fall to the base of the cliff 50.0 m
below. (a) How long does it take the ball to reach the
base of the cliff? (b) What is the total distance
traveled by the ball? Ignore air resistance (likely to
be significant, so our result is an approximation).

The person in Fig stands on the edge of a cliff. The ball falls to the base
of the cliff, 50.0 m below.

Solution: a. We use the same quadratic formula as before, we find t =

5.07 s (the negative solution is physically meaningless).
b. The ball goes up 11.5 m, then down 11.5 m + 50 m, for a total
distance of 73.0 m.
 Summary of Chapter
• Kinematics is the description of how objects move with respect to a
defined reference frame.
• Displacement is the change in position of an object.
• Average speed is the distance traveled divided by the time it took;
average velocity is the displacement divided by the time.
• Instantaneous velocity is the average velocity in the limit as the time
becomes infinitesimally short.
 Summary of Chapter
• Average acceleration is the change in velocity divided by the time.
• Instantaneous acceleration is the average acceleration in the limit as the
time interval becomes infinitesimally small.
• The equations of motion for constant acceleration are given in the text;
there are four, each one of which requires a different set of quantities.
• Objects falling (or having been projected) near the surface of the Earth
experience a gravitational acceleration of 9.80 m/s 2.
Thank You