Acceleration
When an unbalanced force acts on an object it accelerates.
An unbalanced force acting on a stationary object will make it
move.
An unbalanced force can make a moving object get faster or
slower or change direction.
Acceleration is the change in
velocity per unit time.
The acceleration or deceleration of an object is calculated using:
final velocity
(ms-1)
acceleration
(ms-2)
v -u
a
t
initial velocity
(ms-1)
time (s)
Measuring Acceleration
Diagram
Stopwatch
Electronic
Timer
Light Gate
Length of Card
Trolley
Start Line
Finish Line
Method
A trolley is fitted with a length of card. The length of card is
measured using a ruler.
The trolley is released from rest, so the initial velocity is 0 ms-1.
The time taken for the trolley to travel from the start to the finish
line is measured using a stopwatch.
An electronic timer measures the time taken for the length of card to
pass through the light beam. The timer starts when the card breaks
the light beam and stops when the card leaves the light beam.
The final velocity is calculated using:
length of card
v
time
These measurements are used to calculate acceleration in the
relationship shown:
v u
a
t
Results
u  0 ms -1
length of card
v

time
t
a?
s
ms -1
Motion Graphs
There are three types of motion graph we will study:
1. A velocity - time graph.
2. A displacement – time graph.
3. An acceleration – time graph.
displacement-time
velocity-time
acceleration-time
Constant Velocity
s
v
t
a
t
t
Constant Acceleration
s
v
t
a
t
t
Constant Deceleration
s
v
a
t
t
t
Drawing Graphs
Each velocity – time graph has a corresponding acceleration – time
graph and displacement- time graph.
Example
The following velocity – time graph describes a journey.
velocity /
ms-1
12
0
3
10
12
time / s
Draw the corresponding acceleration – time graph.
0-3 seconds
v u
a
t
12  0

3
a  4 ms 2
3-10 seconds
v u
a
t
12  12

7
a  0 ms 2
10-12 seconds
a
v u
t
0  12

2
a  -6 ms 2
acceleration
/ ms-2
4
0
-6
3
10
12
time / s
Q1.
Copy out the following velocity - time graph and underneath it
draw the corresponding acceleration – time graph.
velocity /
ms-1
24
20
0
4
9
13
16
time / s
0-4 seconds
v u
a
t
20  0

4
a  5 ms 2
13-16 seconds
v u
a
t
0  24

3
a  8 ms 2
4-9 seconds
a  0 ms 2
9-13 seconds
v u
a
t

24  20
4
a  1 ms 2
acceleration
/ ms-2
5
1
0
-8
4
9
13
16
time / s
Q2.
Using the following acceleration – time graph of an object
starting from rest, draw the corresponding velocity – time
graph.
acceleration
/ ms-2 5
0
4
-2
v  u  at
 0  5  4
v  20 ms -1
8
18
v  u  at
 20  - 2 10
 20 20
v  0 ms -1
time / s
velocity /
ms-1
20
0
4
8
18
time / s
Q3.
Copy out the following velocity – time graph and underneath it
draw the corresponding acceleration – time graph (after
appropriate calculations).
velocity / ms-1
15
3
0
-9
3
7
9
14
18
24
time / s
0-3 seconds
v u
a
t
3-7 seconds
a  0 ms 2
15  3
3
a  4 ms 2
a  0 ms 2
v u
a
t
3  15
2
a  -6 ms 2

9-14 seconds
7-9 seconds

14-18 seconds
a
v u
t
-9 3
4
a  -3 ms 2

18-24 seconds
v u
t
0  - 9 

6
a
a  1.5 ms 2
acceleration /
ms-2
4
0
-6
3
7
9
14
18
24
time / s
Motion and Direction
In the velocity-time graphs so far, motion has been in only one
direction.
A velocity time graph however can show two different directions;
one direction is positive, while the other is negative.
Example 1
A car travelling along a straight section of road has the following
velocity-time graph.
forward motion
backward motion
(a)
Calculate the distance travelled by the car.
(b)
Calculate the displacement of the car.
(a) Distance travelled is the area under velocity-time graph.
d1 
1
 100  20
2
d2 
 1,000 m
1
 50  10
2
 250 m
d4 
d3  100  10
 1,000 m
1
 50  10
2
 250 m
dtot  2,500 m
(b) Displacement is the forwards motion, less the backward motion.
displaceme nt  1,000 - 250  1,000  250
 500 m
Worksheet – Graphs of Motion
Q1 – Q9
Ball Thrown Into Air
A ball is thrown directly upwards into the air.
It rises into the air and falls back down to the thrower.
The velocity – time graph and corresponding acceleration – time
graph are shown.
velocity / ms-1
time / s
acceleration /
ms-2
time / s
-9.8
Bouncing Ball
(No Energy Loss)
A ball is dropped from a height to the ground.
The ball bounces twice with no energy loss and is then caught.
The velocity - time and acceleration - time graphs are as follows:
max height
velocity /
ms-1
time / s
hits ground
downwards (falling):
- ve
upwards (rising):
+ ve
acceleration /
ms-2
time / s
-9.8
TASK
Re-draw the velocity-time graph for the bouncing ball, taking
downward motion as positive, and upward motion as negative.
hits ground
velocity /
ms-1
time / s
max height
downwards (falling):
+ ve
upwards (rising):
- ve
Bouncing Ball
(With Energy Loss)
A ball is dropped from a height to the ground.
Kinetic energy is lost with each bounce.
The velocity - time and acceleration - time graphs are as follows:
velocity /
ms-1
time / s
acceleration /
ms-2
time / s
-9.8
TASK
Re-draw the velocity-time graph for the bouncing ball, taking
downward motion as positive, and upward motion as negative.
Worksheet – More Graphs of Motion
Q1 – Q7
Equations of Motion
There are three equations of motion.
You must be able to use and derive these three equations of
motion.
1st Equation of Motion
a
v u
t
v u  a t
v ua t
2nd Equation of Motion
To derive the second equation of motion, the velocity-time graph
shown is used as a starting point.
velocity/ms-1
}
v
u
t
v - u
time/s
Displacement (s) is the area under a velocity time graph.
s  area
 area
 ut 
1
t v - u
2
but from equation (1):
v  u  at
v - u  at
so we can rewrite as:
s  ut 
1
t at 
2
s  ut 
1 2
at
2
3rd Equation of Motion
This equation links final velocity (v) and displacement (s).
v  u  at
v2  u  at 
2
v2  u  at u  at 
v2  u2  2uat  a2t2
taking a common factor of 2a gives
1 2

v  u  2a ut  at 
2


2
2
and since
s = ut + ½at2
v 2  u2  2as
Direction
When using the equations of motion, it is essential that direction
is considered.
In these examples, upward motion is taken as positive, so any
downward motion is taken to be negative.
Example
A helicopter is travelling upwards
with a velocity of 25 ms-1.
A package is released and hits the
ground 14 s later.
Path of Package
stationary
( 0 ms-1 )
*
released
*
* hits
ground
This example has motion in two directions.
It is necessary to distinguish between the two directions.
Choose the upward direction as positive!
(a)
How long will it take the package to reach its maximum
height?
t?
u   25 ms 1
v  0 ms 1
a  9.8 ms 2
v  u  at
0  25  - 9.8 t
9.8 t  25
t  2.55 s
(2)
(b)
How high as it climbed since being released?
s?
u   25 ms 1
t  2.55 s
a   9.8 ms 2
(2)
1
s  ut  at2
2
1
 25  2.55  - 9.8  2.552
2


 63.75   31.86
s  31.9 m
(c)
Calculate the velocity of the package just before it hits the
ground.
(2)
v?
u   25 ms 1
t  14 s
a  9.8 ms 2
v  u  at
 25  - 9.814
 25  - 137.2
v  112.2 ms 1
The negative indicates
travelling downwards.
(d)
How high above the ground is the helicopter when the
package is released?
s?
u   25 ms 1
t  14 s
a  -9.8 ms 2
1
s  ut  at2
2
 25  14  
1
- 9.8  14 2
2

 350   960.4
s  610.4 m
So the helicopter is 610.4 m above the ground.
(2)

Worksheet – Equations of Motion
Q1 – Q15
Acceleration Due To Gravity
Diagram
Timer
Ball Bearing
Trap Door
Method
The ball bearing is released from rest, so initial velocity (u) is 0 ms-1.
The displacement (s) of the ball bearing is the distance between the
release point and trap door, and is measured using a metre stick.
An electronic timer measures the length of time taken for the ball
bearing to reach the trap door.
Calculation
u  0 ms -1
s
m
t
a?
s
1
s  ut  at2
2
Projectiles (Half Trajectory)
An object projected sideways through the air will follow a curved
trajectory.
horizontal motion
(steady speed)
VH 
vertical motion
DH
t
accelerates downwards at -9.8 ms-2
The horizontal and vertical motions should be treated separately.
Time is the only quantity common to both.
vh
This is an example of a
‘half-trajectory.’
vv
GREEN – actual motion
RED – vertical motion
BLUE – horizontal motion
At any point in its trajectory, the velocity of a projectile has two
components.
• one vertical, VV
• the other horizontal, VH
The resultant velocity is found drawing a vector diagram and add the
vectors together, TIP to TAIL.
Vector Diagram
horizontal velocity
resultant/actual
velocity
vertical
velocity
Example
A ball is kicked horizontally off an embankment, with a velocity of 30
ms-1.
It lands 24 m from the base of the embankment.
(a)
Calculate how long the ball was in flight.
30 ms-1
24 m
VH 
DH
t
30 
24
t
t
24
30
t  0.8 s
common to
horizontal and
vertical motions
(b)
Calculate the horizontal velocity just before hitting the ground.
Horizontal
Vertical
u  30 ms -1
a  -9.8 ms -2
t  0.8 s
u  0 ms -1
s  24 m
t  0.8 s
a  0 ms -2
travels
horizontally at
steady speed –
no acceleration
horizontally
Horizontal
v  u  at
 30  0  0.8
v  30 ms -1
acted upon by
gravity
not initially
falling down, so
speed of zero in
vertical direction
(c)
Calculate the vertical velocity just before hitting the ground.
Vertical
v  u  at
 0  - 9.8 0.8
v  -7.84 ms -1
(d)
means 7.84 ms-1 downwards
How high is the embankment?
Vertical
s  ut 
a  -9.8 ms -2
t  0.8 s
u  0 ms -1
v  -7.84 ms -1
1 2
at
2
 0  0.8 
s  3.14 m
1
- 9.8  0.82
2


means ball fell through
distance of 3.14 m
so height of the embankment is 3.14 m
(e)
Calculate the resultant velocity of the ball, just before hitting
the ground.
30 ms-1
θ
velocity
-7.8 ms-1
Size
By Pythagoras:
a 2  b2  c 2
resultant velocity 2  302  - 7.842
 900 61.5
resultant velocity  961.5
 31 ms -1
Direction
adj
cos θ 
hyp
30
31
θ  cos 1 0.97
cos θ 
30 ms-1
θ
velocity
-7.8 ms-1
θ  14.6
resultant velocity  31 ms -1 at angle of 14.6 below horizon
Q1.
A ball is kicked off a cliff with a horizontal speed of 16 ms-1.
The ball hits the ground 2.2 s later.
(a)
Calculate the height of the cliff.
(b)
Calculate the distance between the foot of the
35.2 m
cliff and where the ball lands.
(c)
Calculate the vertical component of the balls
velocity just before it hits the ground. 21.6 ms-1
(d)
Calculate the balls velocity as it hits the
ground.
26.9 ms-1 at angle
23.7 m
of 53.5° below
horizon
You may want to draw a diagram to help you get started !!!
Q2.
A ball is kicked off a cliff with a horizontal speed of 22 ms-1.
the ball hits the ground 1.5 s later.
11 m
(a)
Calculate the height of the cliff.
(b)
Calculate the horizontal distance from the foot
33 m
of the cliff, to where the ball lands.
(c)
Calculate the vertical component of the balls
14.7 ms-1
velocity as it hits the ground.
(d)
Calculate the balls actual velocity as it hits the
ground.
26.5 ms-1 at angle of
34° below horizon
You may want to draw a diagram to help you get started !!!
Does Projectile Theory Work?
Diagram
ball-bearing
sand
h
d
Measurements
Horizontal Velocity
measure distance ball-bearing travels
along desk and divide by time taken
Vertical Displacement
measure height of desk from floor
Calculation
Calculate the time of flight.
Vertical
s
m
a  -9.8 ms -2
u  0 ms -1
t?
s  ut 
1 2
at
2
Now calculate the horizontal displacement.
Horizontal
vH 
t
s  vH  t
ms -1
s
s?
Experimentally
The horizontal displacement was measured experimentally using a
metre stick to be
m.
Projectiles (Full Trajectory)
A projectile does not need to be an object falling, but could be an
object fired at angle to the horizontal.
θ
The subsequent motion would be
max height
If air resistance is ignored, the trajectory has an axis of symmetry
about the mid point (maximum height).
So the time taken to reach the maximum height is the same as the
time taken to fall back to the ground.
Various calculations can be made, but firstly, the initial velocity must
be split into its horizontal and vertical components.
Horizontal
Vertical
a = 0 ms-2
a = -9.8 ms-2
Example 1
A golf ball is hit off the tee at 48 ms-1 at angle of 20° to the
horizontal.
48 ms-1
20°
Calculate the horizontal and vertical components of the initial velocity.
Horizontal
adj
cos θ 
hyp
cos 20 
vH
48
48 ms-1
20°
VH
VV
Vertical
sin θ 
sin 20 
opp
hyp
vV
48
vH  48  cos20
vV  48  sin 20
vH  45.1 ms -1
vV  16.4 ms -1
Example 2
An arrow is projected into the air with a velocity of 38 ms-1 at an angle
of 25° to the horizontal.
38 ms-1
250
(a)
Calculate the horizontal and vertical components of the initial
velocity.
Horizontal
adj
cos θ 
hyp
v
cos 25  H
38
38 ms-1
25°
VH
VV
Vertical
adj
hyp
v
sin 25  V
38
sin θ 
vH  38  cos 25
vV  38  sin 25
vH  34.4 ms -1
vV  16.1 ms -1
(b)
Calculate the arrow’s maximum height.
Vertical
a  -9.8 ms -2
u  16.1 ms
v  0 ms -1
s?
-1
v 2  u2  2as
02  16.12  2  -9.8 s
0  259.21  19.6 s
19.6 s  259.21
s  13.2 m
(c)
Calculate the time taken for the arrow to reach its maximum
height.
Vertical
a  -9.8 ms -2
u  16.1 ms
v  0 ms
v  u  at
-1
0  16.1  - 9.8 t
9.8 t  16.1
t  1.64 s
-1
s  13.2 m
t?
(d)
Calculate the total time of the arrows flight.
total time  time up  time down
 1.64  1.64
 3.28 s
(e)
Calculate the horizontal distance travelled by the arrow until
impact with the ground.
Horizontal
u  34.4 ms
a  0 ms -2
t  3.28 s
s?
-1
s  ut 
1 2
at
2
1

 34.4  3.28    0  3.282 
2

s  112.8m
(f)
Calculate the arrow’s velocity 0.5 s after being fired.
Firstly calculate the vertical component of velocity (horizontal
component is constant, since a = 0 ms-2)
Vertical
a  -9.8 ms -2
u  16.1 ms -1
t  0.5 s
v?
v  u  at
 16.1  - 9.8 0.5
 16.1 - 4.9
v  11.2 ms 1
Now calculate the actual velocity after combining the vertical
and horizontal components of the velocity after 0.5 s.
Size
v
By Pythagoras:
θ
a 2  b2  c 2
velocity 
2
 34.4  11.2
 1,183.36  125.44
2
2
34.4 ms-1
velocity  1,308.8
 36.2 ms -1
Direction
tan θ 
opp
adj
11.2
tan θ 
34.4
θ  tan 1 0.326
θ  18
Velocity of the arrow after
0.5 s is:
36.2 ms-1 at angle of 18°
above the horizon
11.2 ms-1
Q1.
A shell is fired from a gun with a velocity of 72 ms-1 at an angle
of 60° to the horizontal.
(a)
Calculate the horizontal and vertical
components of the initial velocity.
VH = 36 ms-1
VV = 62.4 ms-1
(b)
Calculate the maximum height reached.
(c)
Calculate the time taken for the shell to reach
6.4 s
it’s maximum height.
(d)
Calculate the total time of flight.
(e)
Calculate the horizontal range of the shell.
(f)
Calculate the shells velocity after 2.3 s.
199 m
12.8 s
458 m
53.7 ms-1 at angle of 48° above
the horizon
Q2.
to
An arrow is fired with a velocity of 50 ms-1 at an angle of 30°
the ground.
(a)
Calculate the time taken for the arrow to reach
2.55 s
its maximum height.
(b)
Calculate the maximum height reached by the
31.89 m
arrow.
(c)
Calculate the time the arrow is in flight.
(d)
Calculate how far away from the firing point the
arrow will land.
5.1 s
(e)
Calculate the actual velocity of the arrow 1s
after it is fired.
45.89 ms-1 at angle of 19.3°
above horizon
5.1 s