AS-Level Maths:
Mechanics 1
for Edexcel
M1.3 Kinematics
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Contents
Motion graphs
Motion graphs
Formulae for constant acceleration
Examination-style questions
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Kinematics
Kinematics involves the study of how things move.
It is only concerned with the motion itself, not the forces that
cause this motion.
The kinematics of an object is described in terms of its
distance,
displacement,
speed,
velocity,
acceleration.
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Distance and displacement
Distance is a scalar quantity.
The distance a body has travelled is literally the amount of
‘ground’ it has covered during its motion.
Displacement is a vector quantity.
Displacement describes how far a body is from its starting
point and in what direction.
Distance and displacement are measured in metres, m.
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Speed, velocity and acceleration
Speed is a scalar quantity.
The speed of a body relates to how fast the body is travelling.
Velocity is a vector quantity.
The velocity of a body relates to how fast the body is travelling
and in what direction. It is the rate at which a body changes its
position.
Speed and velocity are measured in metres per second, ms–1.
Acceleration can be a scalar or a vector quantity.
Acceleration is the rate of change of speed or velocity.
It is measured in metres per second per second, ms–2.
Negative acceleration is often called deceleration or
retardation.
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Motion graphs
The kinematics of a body can be represented graphically.
The most common graphs are position-time, speed-time,
velocity-time and acceleration-time graphs.
The gradient of a distance-time graph gives speed.
The gradient of a displacement-time graph gives velocity.
The gradient of a velocity-time graph gives acceleration.
The area under a speed-time graph gives the distance
travelled.
The area under a velocity-time graph gives the change in
displacement.
The area under an acceleration-time graph gives the change
in velocity.
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Displacement-time graph
Displacement (m)
This graph shows a journey of 2000 m. It includes a stop of 1
hour after travelling 1000 m metres. The person then returns
to their starting position.
The gradient of this graph
1000
gives velocity.
800
For the first part of the journey,
1000
velocity =
20
= 500 metres per minute
600
400
200
0
0
20 40 60 80 100 120
For the second part of the
journey the velocity is zero.
1000
For the last part, velocity =
30
= –33.3 metres per minute
Time (mins)
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Distance-time graph
This graph also shows a journey of 2000 m with a 1 hour stop.
However, for this graph there is no indication of direction.
The gradient of this graph
gives speed.
Distance (m)
2000
1600
For the first part of the journey,
1000
speed =
20
= 500 metres per minute
1200
800
400
0
0
20 40 60 80 100 120
For the second part of the
journey the speed is zero.
1000
For the last part, speed =
30
= 33.3 metres per minute
Time (mins)
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Velocity-time graph
Velocity (ms–1)
The area under a velocity-time graph gives displacement.
In this example, the area
under the graph is given by a
trapezium with height 12.5
and parallel sides of length
130 and 90.
12.5
10
7.5
5
2.5
0
0
20 40 60 80 100 120 140
Time (s)
Displacement = 21 (130 + 90)×12.5
= 1375 m
The gradient of the graph gives acceleration.
The first part of the graph shows an acceleration of 0.42 ms–2,
the second part 0 and the last part a deceleration of 1.25 ms–2.
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Acceleration-time graph
This graph shows constant acceleration.
Acceleration (ms–2)
6
4
2
0
–2
2
4
6
8
10 12 14
Time (s)
–4
–6
The first part shows an acceleration of 4 ms–2.
The second part shows a deceleration of 6 ms–2.
The last part shows constant velocity.
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Acceleration-time graph
The area under an acceleration-time graph gives change in
velocity.
Acceleration (ms–2)
6
4
2
0
–2
2
4
6
8
10 12 14
Time (s)
–4
–6
For the first part, change in velocity = 4 × 8 = 32 ms–1
For the second part, change in velocity = –6 × 3 = –18 ms–1
There is no change in velocity for the last part.
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Graphs example 1
A man travels in a lift from the top floor of a hotel to reception
on the ground floor.
The lift accelerates with a constant acceleration of 1 ms–2
until it reaches a constant velocity of 4 ms–1.
It then travels at this constant velocity for t seconds before
decelerating with a constant deceleration of 2 ms–2 until it
reaches the ground floor.
Given that the man has descended 44 m,
a) sketch the velocity-time graph of the lift and use it to find t
b) sketch the acceleration-time graph of the lift.
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Graphs solution 1
The distance travelled is given
by the area under the graph, so
4
( 21 × 4 × 4) + (4t ) + ( 21 × 4 × 4) = 44
4
t
Time (s)
The acceleration-time
graph for the lift can
then be sketched as
follows:
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8 + 4t + 4 = 44
4t = 32
t = 8 secs
2
Acceleration (ms–2)
Velocity (ms–1)
The velocity-time graph for the lift can be sketched as follows:
1
0
2
4
6
8 10 12 14
Time (s)
–1
–2
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Graphs example 2
A car and a motorcycle are travelling along a straight road.
The car accelerates from rest to a constant speed of 28 ms–1.
The motorcycle accelerates from rest to a constant speed of
25 ms–1 in 10 seconds.
After travelling for 90 seconds the car hits traffic and
decelerates to a constant speed of 22 ms–1 in 5 seconds.
The motorcycle is unaffected by the traffic and maintains his
speed.
The motorcycle overtakes the car after they have both
travelled 3700 m.
Draw a speed-time graph and use it to find the time when the
motorcycle overtakes the car and how long the car was initially
accelerating for.
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Graphs solution 2
Velocity (ms–1)
t1
28
25
22
0
Motorbike
Car
10
90 95
Time (s)
Let t1 be the time for which the motorbike is travelling with a
constant speed before it overtakes the car.
The motorbike overtakes the car after travelling 3700 m so,
( 21 ×10 × 25) + (t1 × 25) = 3700
125 + 25t1 = 3700
25t1 = 3575
t1 = 143 seconds
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Graphs solution 2
So the motorbike accelerates for 10 seconds and then travels
at a constant speed for 143 seconds before overtaking the car.
 The motorbike travels for 153 seconds before overtaking the
car.
Let t2 be the time for which the car is initially accelerating.
Velocity (ms–1)
t2
28
25
22
0
Motorbike
Car
10
90 95
153
Time (s)
This area represents 3700 m
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Graphs solution 2
Since the area under the graph for the car between 0 and 153
seconds is equal to 3700 so we can write,
( 21 × t2 × 28) + ((90  t2 )× 28) + ( 21 (22 + 28)× 5) + (58 × 22) = 3700
14t2 + 2520 – 28t2 + 125 + 1276 = 3700
14t2 = 221
t2 = 15.8 (to 3 sf)
Therefore the car was initially accelerating for 15.8 s (to 3 s.f.)
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Contents
Formulae for constant acceleration
Motion graphs
Formulae for constant acceleration
Examination-style questions
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Formulae for constant acceleration
If a particle is moving in a straight line with a constant
acceleration then there are five equations of motion that can
be used to determine missing quantities.
v = u + at
s = 21 (u + v )t
s = ut + 21 at 2
s = vt  21 at 2
v 2 = u 2 + 2as
Where
s = displacement in metres
u = initial velocity in ms–1
v = final velocity in ms–1
a = acceleration in ms–2
t = time taken in seconds
These are sometimes called the
suvat formulae.
For vertical motion acceleration due to gravity is g, 9.8ms–2.
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v = u + at
The motion of an object with initial velocity u and final velocity
v over time t can be illustrated using a velocity-time graph.
Velocity (ms–1)
By definition, acceleration is
the rate of change of velocity.
v
The constant acceleration a is
therefore given by the gradient
of the graph. So
u
vu
t
at = v – u
a=
t
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Time (s)
v = u + at
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s = ½(u + v)t
We can use the same graph to find the distance s travelled by
an object with initial velocity u and final velocity v over time t.
Velocity (ms–1)
This distance is given by the
area under the graph.
v
This area is a trapezium with
parallel sides of length u and v
and width t. So
u
s = 21 (u + v )t
This can also be written as
t
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Time (s)
u+v
s=
t

 2 
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s = ut + ½at2
distance travelled = area of rectangle A + area of triangle B
= ut + 21 t (v  u )
vu
a=
so at = v – u
t
This gives us distance travelled = ut + 21 t ( at )
So
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s = ut + 21 at 2
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s = vt – ½at2
distance travelled = area of rectangle C – area of triangle D
= vt  21 t (v  u )
We have shown that at = v – u
This gives us
So
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distance travelled = vt  21 t ( at )
s = vt  21 at 2
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v2 = u2 + 2as
We can show that v2 = u2 + 2as as using
1
v = u + at
u+v
s=
t
2

 2 
Rearranging equation 1 to make t the subject gives
(v  u )
t=
a
Substituting this into equation 2
 u + v  v  u 
s=


 2  a 
2as = (u + v)(v  u )
2as = v2  u 2
So
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v 2 = u 2 + 2as
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Constant acceleration example 1
A stone is thrown vertically upwards with a speed of 10 ms–1
from a point 15 m above the ground. Find the maximum
height above the ground that the stone reaches and find the
time taken for the stone to reach the ground.
Taking  to be positive, the
information given in the question is:
u = 10
a = –g
(Take g to be 9.8)
The question is asking for s when
v = 0, and for t when s = –15.
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Constant acceleration solution 1
To calculate s when v = 0, given u and a requires the use of
v2 = u2 + 2as.
02 = 102 + 2(–9.8)(s) 
0 = 100 – 19.6s
19.6s = 100

s = 5.10 (to 3 s.f.)
Therefore, the maximum height above the ground that the
stone reaches is 20.1 m (to 3 s.f.).
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Constant acceleration solution 1
To calculate t when s = –15, given u and a requires the use of
s = ut + 21 at2.
–15 = 10t + 21 (–9.8)t2 
–15 = 10t – 4.9t2
Arranging all the terms on the left gives us the following
quadratic equation
4.9t2 –10t – 15 = 0
b  b 2  4ac
Using
gives the solution
2a
t = 3.05 or t = –1.01 (to 3 s.f.)
Therefore the stone reaches the ground after 3.05 s (to 3 s.f.).
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Constant acceleration example 2
A particle moves in a horizontal line from a point A to a point
C, via point B. It has a constant acceleration of 1 ms–2 and
passes point B after 6 seconds and point C after a further 4
seconds. Its velocity at C is 50 ms–1. Calculate the velocity at
A and the distances AB and BC.
We can sketch the situation as follows
A
B
t=6
C
t = 10
v = 50
Taking  to be positive the question firstly asks for u when
a = 1, t = 10 and v = 50. This requires the use of v = u + at.
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Constant acceleration example 2
v = u + at
50 = u + (1)(10)
50 = u + 10
 u = 40
Therefore the particle passes A with a velocity of 40 ms–1.
The next part of the question asks for s when u = 40, a = 1 and
t = 6, requiring the use of s = ut + 21 at2.
s = ut + 21 at2
s = 40(6) + 21 (1)(6)2
s = 240 + 18 = 258
Therefore AB is 258m.
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Constant acceleration example 2
The final part of the question asks for s when u = 40, a = 1
and t = 10, again requiring the use of s = ut + 21 at2.
s = ut + 21 at2
s = 40(10) + 21 (1)(10)2
s = 400 + 50 = 450
Therefore AC is 450 m and so BC is 192 m.
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Constant acceleration example 3
A ball falls off a cliff and lands on the beach 3.2 seconds later.
How high is the cliff?
Taking  to be positive the information given in the question
is
a = 9.8
t = 3.2
u=0
To calculate s requires the use of the formula s = ut + 21at2.
s = ut + 21 at2
s = (0)(3.2) + 21 (9.8)(3.2)2
s = 50.2 (to 3 s.f )
Therefore the height of the cliff is 50.2 m (to 3 s.f.)
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Contents
Examination-style questions
Motion graphs
Formulae for constant acceleration
Examination-style questions
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Examination-style question 1
A ball is thrown vertically upwards with an initial velocity of
ms–1 from a point 1.2 m above the ground. It reaches a
maximum height of 23 m above the ground.
Calculate
a) the initial velocity
b) the velocity with which the ball strikes the ground
c) the total time the ball is in the air.
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u
Solution 1
a) Taking  as positive, v = 0, s = 21.8, a = –9.8
v2 = u2 + 2as
0 = u2 + 2(-9.8)(21.8)
0 = u2 – 427.28
u = ±427.28 (to 3 s.f.)  initial velocity is 20.7ms–1 (to 3 s.f.)
b) Method 1 - Using downward motion only
Taking  to be positive, u = 0, a = 9.8, s = 23
v2 = u2 + 2as
v2 = 02 + 2(9.8)(23)
v2 = 450.8
v = ± 21.2 (to 3 s.f.)
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 The ball hits ground with a
velocity of 21.2 ms–1 (to 3 s.f.)
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Solution 1
b) Method 2 – Using whole motion
Taking  to be positive, u = –427.28, a = 9.8, s = 1.2
v2 = u2 + 2as
v2 = 427.28 + 2(9.8)(1.2)
v2 = 450.8
v =  21.2 (to 3 s.f.)
 ball strikes ground with a velocity
of 21.2 ms-1 (to 3 s.f.)
c) Taking  as positive, u = –427.28, v = 450.8, a = 9.8
v = u + at
(v  u )
t=
a
( 450.8  427.28 )
t=
9.8
t = 4.28 (to 3 s.f.)  ball is in the air for 4.28 s (to 3 s.f.)
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Examination-style question 2
A car is travelling with a uniform acceleration along a
straight road.
It passes a point A with a velocity of 8 ms–1 and 5 seconds
later it passes a point B with velocity 25 ms–1.
Find the velocity with which the car passes the mid-point of
AB.
The distance AB needs to be found first: u = 8, v = 25, t = 5
s = 21 (u + v)t
s = 21 (8 + 33)(5)
s = 82.5
 the distance AB is 82.5 m and so the mid-point of AB is at a
distance of 41.25 m from A.
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Solution 2
Acceleration now needs to be found: u = 8, v = 25, t = 5
 a=
v = u + at
a=
(v  u )
t
(25  8)
5
a = 3.4
 the acceleration of the car is 3.4 ms–2.
The velocity of the car as it passes the mid-point of AB can
now be found: u = 8, a = 3.4, s = 41.25
v2 = u2 + 2as
v2 = 82 + 2(3.4)(41.25) = 344.5
v = ±344.5  the car passes the mid-point of AB with
a velocity of 18.6 ms–1 (to 3 s.f.)
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