Engineering Sciences

BPS1153
Linear Motion
By: Dr Wan Norlinda Roshana Mohd Nawi
Content
• Introduction to acceleration
• Free fall
• Projectiles

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Learning Outcomes
• Define acceleration and state its units
• Apply that ‘free fall’ has a constant acceleration of 9.8 m/s2
• Use the equation of motion v = u + at in calculations

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Introduction to acceleration

v v f − vi
a = acceleration = = • Change in velocity over time.
t t
• Either hitting the gas or hitting the
• →delta.
break counts as acceleration.
• Means “change in”
• Units are m/s2
and is calculated by
subtracting the initial
value from the final
value.

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Acceleration – how fast you speed up, slow down, or change
direction; it’s the rate at which velocity changes. Two examples:

t (s) v (mph) t (s) v (m/s)

0 55 0 34
1 57 1 31
2 59 2 28
3 61 3 25

a = +2 mph / s m/s
a = -3 s = -3 m / s 2
VELOCITY
ACCELERATION + Speeding up
_ Slowing down
Linear Motion Equation

v f = vi + at s = 1 (vi + v f )t
2

s = vi t + at
1
2
2
v = v + 2as
2
f
2
i

Always assume that acceleration is constant.

Always use instantaneous velocities, not average velocities
Hidden variable: Starting from rest corresponds to a vi=0
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Derivations

a = v / t (by definition)
a = (vf – vi) / t
 vf = vi + a t

s = ½ (vi + vf) t = ½ (vi + vi + a t) t

 s= vi t + a t 2 1
2

(cont.)
 Derivations
vf = vi + a t  t = (vf – vi) / a
1
s = vi t + 2
at2
 s=vi [(vf – vi) / a] + 1
a [(vf – vi) / a] 2
2
 vf2 – vi2 = 2 a s
Note that the top equation is solved for t and that expression for t is
substituted twice (in red) into the s equation. You should work out the
algebra to prove the final result on the last line.
Example

A car going 15 m/s accelerates at 5 m/s2 for

3.8s. How fast is it going at the end of the
acceleration?

First step is identifying the variables in the equation

and listing them.
Example: Solution

A car going 15 m/s accelerates at 5 m/s2 for

3.8 s. How fast is it going at the end of the
acceleration?

t=3.8s
vi=15m/s
a=5m/s2
vf=?
Answer
• vf = vi + a t
• Vf = 15 m/s + (5 m/s2 x 3.8 s)
• = 34 m/s
EXERCISE
1. A coach increases velocity from 4 km/h to 40 km/h at an average
acceleration of 0.2 m/s2. Find the time taken for this increase in
velocity.

2. A ship changes velocity from 15 km/h to 20 km/h in 25 min.

Determine the average acceleration in m/s2 of the ship during this
time.
EXERCISE
A typical jetliner lands at a speed of 260 km/h and decelerates at the
rate of 20 m/s2. If the plane travels at a constant speed of 260 km/h for
1.00 s after landing before applying the brakes, what is the total
displacement (in meter) of the aircraft between touchdown on the
runway and coming to rest?
 Free Fall
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.

In the absence of air resistance, it is

found that all bodies (large and
small) above the earth fall vertically
with the same acceleration due to
the force of gravity.
 FREELY FALLING BODIES
• If a dense object such as a stone is dropped from a height,
called free-fall, it has a constant acceleration of
approximately 9.8 m/s2.

• In a vacuum, all objects have this same constant

acceleration, vertically downwards, that is, a feather has the
same acceleration as a stone.

• However, if free-fall takes place in air, dense objects have the

constant acceleration of 9.8 m/s2 over short distances, but
objects which have a low density, such as feathers, have little
or no acceleration.
Falling Objects

In the absence of air resistance,

all objects fall with the same
acceleration, although this may
be hard to tell by testing in an
environment where there is air
resistance.
Heavy and light objects fall at the same rate
 Freely Falling Bodies

In the absence of air resistance, it is found that all bodies (large and small)
above the earth fall vertically with the same acceleration due to the force of
gravity.
Furthermore, if the distance of the fall is small compared to the radius of the
earth, the acceleration remains essentially constant throughout the fall.
This idealized motion, in which air resistance is neglected and the acceleration
is nearly constant, is known as free-fall.
Acceleration Due to Gravity

The acceleration of a freely falling body is called the acceleration

due to gravity, g.

The acceleration due to gravity is directed

downward, toward the center of the earth.
Near the earth's surface, g = 9.80 m/s2
Falling Objects

The acceleration due to

gravity at the Earth’s
surface is approximately
9.80 m/s2.
Equation of Motion
Since the acceleration due to gravity is constant in the free-fall, we can
use all the equations of kinematics from the previous topic

v f = vi + at s = 1 (vi + v f )t
2

s = vi t + at
1
2
2
v = v + 2as
2
f
2
i
If an object is projected
upwards in a perfectly
vertical direction, then the
velocity at which it is
projected is equal in
magnitude and opposite in
sign to the velocity that it
has when it returns to the
same height. That is, a ball
projected vertically with an
upward velocity of +30 m/s
will have a downward
velocity of -30 m/s when it
returns to the same height
Example
1) A stone is dropped from an airplane.
Determine
(a) its velocity after 2 s and
(b) the increase in velocity during the third second, in
the absence of all forces except that due to gravity.
t=0

t=2

t=5
t=0

t=2

t=5
2) Determine how long it takes an object, which is free-
falling, to change its speed from 100 km/h to 150 km/h,
assuming all other forces, except that due to gravity, are
neglected.

List down all the info,

a = 9.8 m/s2

u = 100 km/hr v f = vi + at
v = 150 km/hr
A Falling Stone
A stone is dropped from rest from the top of a tall building, as the figure indicates.

After 3.00 s of free-fall,

a. what is the velocity v of the stone? (ans: 29.4 m/s)
b. what is the distance of the stone from the roof? (ans: 44.1 m)
c. the velocity during the 4th sec? (ans: 39.2 m/s)
Coin Toss
A football game customarily begins with a coin toss to determine
who kicks off. The referee tosses the coin up with an initial speed
of 6.00 m/s. In the absence of air resistance, how high does the coin
go above its point of release?
Example 1

A ball is thrown vertically from 12

m level in elevator shaft with initial
velocity of 18 m/s. At same
instant, open-platform elevator
passes 5 m level moving upward at
2 m/s.
Determine (a) when and where ball
hits elevator and (b) relative
velocity of ball and elevator at
contact.
SOLUTION:
•Substitute initial position and
velocity and constant acceleration of
ball into general equations for
uniformly accelerated rectilinear
motion.
•Substitute initial position and
constant velocity of elevator into
equation for uniform rectilinear
motion.

•Write equation for relative

position of ball with respect to
elevator and solve for zero
relative position, i.e., impact.
•Substitute impact time into
equation for position of elevator
and relative velocity of ball with
respect to elevator.
SOLUTION:
• Substitute initial position and velocity and constant
acceleration of ball into general equations for
m  m
B =
vuniformly
v 0 + at = 18 −
accelerated t motion.
rectilinear
9.81
s  s2 
 m  m 2
yB = y0 + v0t + 12 at 2 = 12 m + 18 t −  4.905 2 t
 s  s 

• Substitute initial position and constant velocity of

elevator into equation for uniform rectilinear
motion. m
vE = 2
s
 m
y E = y0 + v E t = 5 m +  2 t
 s
 •Write equation for relative position of ball with
respect to elevator and solve for zero relative
position, i.e., impact.

yB E ( )
= 12 + 18t − 4.905t 2 − (5 + 2t ) = 0

t = −0.39 s (meaningless )
t = 3.65 s

•Substitute impact time into equations for position of elevator and

relative velocity of ball with respect to elevator.

y E = 5 + 2(3.65) y E = 12.3 m

v B E = (18 − 9.81t ) − 2 v B E = −19.81

m
s
= 16 − 9.81(3.65)
Projectile Motion
Projectile motion is a special case of two-dimensional motion. A
particle moving in a vertical plane with an initial velocity and
experiencing a free-fall (downward) acceleration, displays projectile
motion.
Projectile motion (2 dimensions)

Note: if we take y to be positive

upward, the ay = - g but if we take y
to be positive downward, ay = g

At t=0, x=0, following Galileo’s idea, the velocity is separated into

horizontal component and vertical component. For this situation it has only
horizontal component for velocity, thus vy is equal to 0 and experiences a
vertically downward acceleration which is force of gravity (ay=g=9.8 m/s2).
In the horizontal direction, there is no acceleration (ax=0), so horizontal
component of velocity remains constant, vxo=vx=constant
For this situation it has both horizontal and
vertical component for velocity

Kinematic
equations for
constant
acceleration
A projectile is fired from the edge of a 150-m cliff with an
initial velocity of 180 m/s at an angle of 30° with the
horizontal. Neglecting air resistance, find (a) the horizontal
distance from the gun to the point where the projectile strikes
the ground, (b) the greatest elevation above the ground
reached by the projectile.
**recap
**recap
Vertical Motion. Uniformly
Accelerated Motion. Choosing the
positive sense of the y axis upward
and placing the origin O at the
gun, we have
 Horizontal Motion. Uniform
Motion. Choosing the positive
sense of the X axis to the right,
we have

Substituting into the equation of uniform motion, we obtain

Horizontal Distance.

When the projectile strikes the ground, we have

Substituting into the equations of
uniformly accelerated motion, we have

Equation 1

Equation 2

Equation 3
 Carrying this value into Eq. (2) for the vertical motion, we
write

Carrying t= 19.91 s into Eq. (4) for the horizontal motion, we

obtain

Greatest Elevation. When the projectile reaches its greatest

elevation, we have vy= 0; carrying this value into Eq. (3) for the
vertical motion, we write
 Example 4

Emanuel Zacchini, the famous human cannonball of the

Ringling Bros. And Barnum & Bailey Circus, was fired out of a
cannon with a speed of 24.0 m/s at an angle of 40.0° to the
horizontal. If he landed in a net 56.6 m away at the same height
from which he was fired, how long was Zacchini in the air?
Solution: Because Zacchini was in the air for the same amount
of time vertically that he was horizontally, you can find his
horizontal time and this will be the answer. First, you need the
horizontal velocity component.
Now you have the horizontal velocity component and the
horizontal displacement, so you can find the time.
End
Thank You

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