PHY433

CHAPTER 3 (a):
MOTION IN 1-D
 Learning Content

3.1 Displacement, velocity and acceleration

3.2 Equation of motion for constant acceleration
3.3 Velocity-time graph
 3.1 Displacement, Velocity &
Acceleration
Reference Frames and Displacement
Any measurement of position, distance, or speed must be made with respect to
a reference frame.

For example, if you are sitting on a train and someone

walks down the aisle, the person’s speed with respect to
the train is a few miles per hour, at most. The person’s
speed with respect to the ground is much higher.
 Displacement & Distance
 Displacement (blue line):
◦ The shortest distance from the initial
to final position of the object.
◦ Unit: meter (m)
◦ Quantity: Vector

 Distance (black line):

 The actual path travel from initial to final position of the
object.
 Unit: meter (m)
 Quantity: Scalar
The displacement is written:

Left: Right:
Displacement is positive. Displacement is negative.
 Average Speed & Velocity
 Average Speed, v
◦ the rate of change of distance.
◦ Unit: meter per second (ms-1)
◦ Quantity: Scalar

 Average velocity, v
◦ the rate of change of displacement.
◦ Unit: meter per second (ms-1)
◦ Quantity: Vector
 Average Speed & Velocity
Example 3.1
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?
 Average Speed & Velocity
Example 3.2
How far can a cyclist travel in 2.5 h along a straight road
if her average velocity is 18 km/h?
 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.
Instantaneous Velocity
 Example 3.3
Given x as a function of t.
A jet engine moves along an
experimental track (which we call the
x axis) as shown. We will treat the
engine as if it were a particle. Its
position as a function of time is given
by the equation x = At2 + B, where A
= 2.10 m/s2 and B = 2.80 m. (a)
Determine the displacement of the
engine during the time interval from t1
= 3.00 s to t2 = 5.00 s. (b) Determine
the average velocity during this time
interval. (c) Determine the magnitude
of the instantaneous velocity at t =
5.00 s.
 Acceleration
Acceleration is the rate of change of velocity.

Example 3.4
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?
 Acceleration

 Negative acceleration – denotes slowing down

process or braking.
 Negative acceleration also known as deceleration.
 Zero acceleration, a = 0 means there is no change in
the velocity or constant velocity (or it could also mean
object is not moving).
 Example 3.5
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?
 Example 3.6
(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?
 Instantaneous Acceleration
The instantaneous acceleration is the average acceleration
in the limit as the time interval becomes infinitesimally short.
 Example 3.7
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.
3.2 Motion at constant acceleration
Kinematic Equations
 Equation of kinematics for constant acceleration

 apply for motion in a straight line

 Kinematic Equations
From the definition of average acceleration, uniform
(constant) acceleration is given by

Uniform acceleration, a  change in velocity

time taken

vu
a t

v  u  at (1)
where v : final velocity
u : initial
velocity
a : 19

uniform
 Kinematic Equations
 From equation (1), the velocity-time graph is shown in
figure:

 From the graph,

The displacement after time, s = shaded area under the
graph
 Hence, =
1
s  u  v  t (2)
the area of trapezium
2
20
 Kinematic Equations

 By substituting eq. (1) into eq. (2) thus:-

1 1 2
s  u  u  a t t s  ut  a t
2 (3)
2
 From eq. v  u
 2s 
(1), at 2s v  u (v  u )   at 
 From eq. (2), v   t 
t
u

v2  u 2  (4)

2as 21
 Example 3.8
Acceleration of a car.
How long does it take a car to cross a 30.0-m-wide
intersection after the light turns green, if the car accelerates
from rest at a constant 2.00 m/s2?
 3.3 Free Falling Object
Near the surface of the Earth, all objects experience
approximately the same acceleration due to gravity.
The position of the
free-falling object at
regular time
intervals, every 1
second, is shown.
The fact that the
distance which the
ball travels every
interval of time is
increasing is a sure
sign that the ball is
speeding up as it
falls downward.
In the absence of air
resistance, all objects
fall with the same
acceleration, although
this may be tricky to tell
by testing in an
environment where
there is air resistance.
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.
• Free falling has no external force except the
gravitational pull.
 Assume a = g = 9.81 m/s2 in the downward direction.
 The acceleration due to gravity is positive (falls).

• There are four (4) cases involved in free falling

objects:
1. Dropped – released from rest
2. Thrown downward
3. Thrown upward
4. Thrown upward then fall back to the ground
• Table below shows the equations of linear motion and free
falling bodies.

Linear motion Freely falling bodies

v  u  at v  u  gt
v2  u 2  2as v2  u 2  2gs
1 2 1 2
s  ut  at s  ut  gt
2 2

If the object is thrown upwards, a = -g

 Case 1: Dropped – released from rest
• Initial velocity is zero.
u = 0 ms-1
• Use the kinematic equations:
– Generally use y instead of x a=g
since vertical.
• Acceleration, ay = g = 9.81 m/s2

Velocity – Time Graph Acceleration – Time

Graph
Velocity, Acceleration,
ms-1 ms-2

0 Time, s 0 Time, s
 Case 2: Thrown downward
• Initial velocity  0 ms-1
u≠0
– With downward being positive, ms-1
initial velocity will be positive.
a=g
• ay = g = 9.81 m/s2

Velocity – Time Graph Acceleration – Time

Graph
Velocity, Acceleration,
ms-1 ms-2

0 Time, s 0 Time, s
 Case 3: Thrown upward
• Initial velocity  0 ms-1
– With acceleration of downward u ≠ 0 ms-1
being positive, moving upward will a=-g
be negative.
• ay = - g = - 9.81 m/s2
Velocity – Time Graph Acceleration – Time
Graph
Velocity, Acceleration,
ms-1 ms-2

0 Time, s

0 Time, s
 Case 4: Thrown upward then fall back to the
ground
v=0
• Initial velocity is upward, so positive.
• The instantaneous velocity at the u≠0
maximum height is zero.
Set:
• ay = - g = - 9.81 m/s everywhere in
2
• a=-g
the motion. • s = y = -ve

Velocity – Time Graph Acceleration – Time

Graph
Velocity, Acceleration,
ms-1 ms-2

0 Time, s
0 Time, s
 SIGN CONVENTION:
A BALL THROWN VERTICALLY
av
= -=
y=0
+ UPWARD

a=-
vy==++ yav =
= -+
=- • Displacement is positive
(+) or negative (-) based
UP = + on LOCATION.

Release Point vay =

==--0
• Velocity is positive (+)
or negative (-) based
on direction of motion.

• Acceleration value is always

yav=
=-
=-
negative
Negative
Negative
 Example 3.9
A ball is thrown vertically upwards with an initial speed of 15 ms-1. Find
the velocity and the height of the ball after 2 s.

Solution:

Using the equation: v  u  gt , g negative because going upwards.

v  u  gt
 (15 ms 1 )  (9.81 ms 2 )( 2 s )  4.6 ms 1

Using the equation: 1 2

s  ut  at
2
1
s  (15 ms )( 2 s )  (9.81 ms  2 )( 2 s ) 2
1

2
 10.4 m
Example 3.10
A person throws a ball upward
into the air with an initial velocity
of 15.0 m/s. Calculate (a) how
high it goes, (b) how long the ball
is in the air before it comes back
to the hand. Ignore air resistance
and (c) the velocity of the ball
when it returns to the thrower’s
hand (point C).
Example 3.11

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).
 Exercise 3.1
1. A man jumped down vertically from the roof of a building
with initial velocity 10 m/s. Height of the building is 72
m. Calculate the time taken for him to reach the ground.
(Ans.: 2.95 s)
2. A ball is thrown vertically upward with initial velocity 20
m/s. What is the highest point of the ball? (Ans.: 20.39 m)
3. A ball is thrown vertically downward with initial velocity
14 m/s from a 10 m building. The velocity of the ball
right before it strikes the ground is 20 m/s. Calculate the
time taken to reach the ground. (Ans.: 0.59 s)
4. A ball is thrown vertically upward from the edge of a 98
m building & reaches the ground 6 s after leaving the
thrower’s hand. Assume that the thrower’s hand is 2 m
above the roof of the building. Calculate the:
a) Initial velocity of the ball. (Ans.: 12.76 m/s)
b) time taken to reach the maximum height. (Ans.: 1.30 s)
c) maximum height (highest point) from the roof of the
building. (Ans.: 10.30 m)
d) velocity of the ball when it strikes the ground. (Ans.:
46.1 m/s)
e) position and velocity of the ball at 1 s and 5 s after
leaving the thrower’s hand. (Ans.: 2.95 m/s & 7.86
m; - 36.29 m/s & - 58.83 m)
5. An object is thrown upwards with a velocity of 30 m/s
from a height of 20 m. Calculate:

a. The maximum height. (Ans.: 45.87 m)

b. Time taken for object to move up. (Ans.: 3.06 s)
c. The time for object to reach initial position. (Ans.:
6.12 s)
3.4 Velocity-Time Graph

Graphical Method
 Example 3.12

A car accelerates at 2 m/s2 from rest. When it reaches a

maximum speed of 40 m/s, it immediately decelerates at a
constant rate for 10 s until it stops.
a. Draw a velocity – time graph for the car’s motion.
b. Determine the total distance covered by the car.