KINEMATICS IN DIRECTION WITH CONSTANT ACCELERATION
1. The ‘suvat’ equations
Consider a point mass (particle) moving with a velocity of 5ms-1 and accelerating to a speed of 20ms-1 at
a constant rate over a 10 second time period.
Draw a velocity/time graph for the particle.
20
5
0
10
What is the acceleration of the particle? 15 ÷ 10 = 1.5ms-2
What is the displacement of the particle after 10 seconds? 1/2(5+20)×10 = 125m
You can see that the graph drawn will be a trapezium and that if we generalise the situation so that the
initial velocity is u and the final velocity is v, the graph drawn will look like this:
velocity
v
u
s
time
t
You can see that the acceleration, a, which is the gradient of the line is given by a 
vu
t
which
rearranges to be v = u + at
You can also see that the displacement, s, which is the area under the graph is given by s 
Combining these equations together it is possible to eliminate t from the second equation:
(u  v )t
2
(u  v)( v  u )
2a
2
u  v2
s
2a
2as  u 2  v 2
s
v 2  u 2  2as
Combining the original equations together to eliminate v from the second equation
(u u  at )t
2
(2ut  at 2 )
s
2
1
s  ut  2 at 2
s
These four equations are sometimes called the ‘suvat’ equations. When you have 3 of the 4 letters in an
equation you can use them to find the fourth. Three of the equations will suffice to solve any problem.
Note, however that they can only be used to model situations involving constant acceleration.
The three most useful equations are:
v  u  at
v 2  u 2  2as
s  ut  12 at 2
2. Using the Equations of Uniform Acceleration
In any problem, you must first write down what you know, and what you want to find out. Then select the
appropriate equation. Each equation connects four of the five quantities u, v, a, t and s.
Example 1 : A car accelerates from rest at 5ms-2. How long does it take for the car to travel 100m,
and what is its velocity at this instant?
NOTE: FOR THIS EXAMPLE IT IS GOOD TO SKETCH A VELOCITY TIME
GRAPH TO SEE HOW THE ANSWERS OBTAINED FROM THE SUVAT
EQUATIONS TIE UP WITH A ‘COMMON SENSE’ APPROACH.
u=0
a=5
s = 100
Find t.
s  ut  12 at 2
100  0  12  5  t 2
t 2  40
t  40
 6.32 seconds (2 d.p.)
u=0
a=5
s = 100
Find v.
v 2  u 2  2as
v 2  0  2  5  100
v 2  1000
v  1000
 31.62 ms -1 (2 d.p.)
Example 2 : A bus leaves a bus stop by accelerating at 0.8ms-2 for 5s. It then travels at a constant
speed for 2 minutes before slowing down uniformly at 4ms -2 to come to rest at the next
bus stop.
a) Find the constant speed in the middle part of the journey
b) Find the distance travelled in the first part of the journey
c) Find the total distance between the two bus stops
NOTE: FOR THIS EXAMPLE IT IS GOOD TO SKETCH A VELOCITY TIME
GRAPH TO SEE HOW THE ANSWERS OBTAINED FROM THE SUVAT
EQUATIONS TIE UP WITH A ‘COMMON SENSE’ APPROACH.
a) u = 0, t = 5, a = 0.8
v = u + at
v = 0 + 5×0.8
v = 4 ms-1
b) u = 0, t = 5, a = 0.8
s = ut + 1/2at2
s = 0 + 1/2 × 0.8× 25
s = 10m
c) Distance in acceleration phase = 10m
Distance in middle section
= 120 × 4 = 280m
Distance in deceleration phase
u = 4, v = 0, a = -4,
v2 = u2 + 2as
0 = 16 – 2 × 4 ×0.8 s
s = 25m
Total distance = 10 + 280 + 25 = 315m
Example 3 : A particle positioned at O is travelling at 20ms-1.
It starts to decelerate at a constant rate of 3ms-2 and continues to do so for 12 seconds.
a) Find the time taken for the particle to come instantaneously to rest.
b) Find the final velocity of the particle.
c) Find the position of the particle relative to O at the end of the journey.
d) Find the distance travelled in the entire 12 seconds
a) u = 20, a = -3, v = 0
v = u + at
0 = 20 -3t
t = 62/3 s
b) u = 20, a = -3, t = 12
v = u + at
v = 20 + -3 × 12
v = -16 ms-1 (speed is 16 in the opposite direction)
c) u = 20, a = -3, t = 12
s = ut + ½at2
s = 20 × 12 + ½× -3 × 144
s = 240 – 216
s = 24 m (so 24 to ‘the right’ of the start position)
e) Distance travelled = Distance up to stopping (at 62/3 s) + Distance back
Distance up to stopping
u = 20, a = -3, v = 0
v2 = u2 + 2as
0 = 400 – 6s
s = 662/3 m
Distance back = 662/3 – 24 = 422/3 m
Total distance = 662/3 + 422/3 = 1091/3 m
(May be worth while showing that had the particle returned to the start position its speed
would have been back up to 20ms-1 but in the negative direction)
Example 4 : A particle is moving in a straight line from O to A with a constant acceleration of 2ms2. Its velocity at A is 30ms-1 and it takes 12 seconds to travel from O to A. Find the
particle’s velocity at O and the distance OA.
a = 2 v = 30 t = 12 Find u and s.
v  u  at
30  u  2 12
u  6 ms -1
The velocity at O is 6ms-1.
s  ut  12 at 2
 6 12  12  2 122
 216 m
The distance OA is 216m.
Example 5 : A train starts from rest at a station S and moves with constant acceleration. It passes a
signal box B 15 seconds later with a speed of 81kmh-1. Find the acceleration of the train
in ms-2 and the distance in metres between the station and the signal box.
We must first change the units of the speed at B into ms-1 :
81 kmh -1  80000 mh -1  22.5 ms -1 .
u = 0 v = 22.5 t = 15 Find a and s.
v  u  at
22.5  0  15a
a  1.5 ms -2
The acceleration is 1.5ms-2.
s  12 (u  v)t
 12 (0  22.5) 15
 168.75 m
The distance between the station and the signal box is 168.75m.
Example 6 : A particle starts from rest and moves along a straight line with a constant acceleration until it
reaches a velocity of 15ms-1. It is then brought to rest again by a constant retardation of 3ms2
. If the particle is then 60m from its starting point, find the time for which the particle is
moving.
Acceleration phase : u = 0 v = 15
Deceleration phase : u = 15 v = 0 a = −3 (negative since the particle is retarding)
There is not enough information on the acceleration phase to proceed, so we consider the
deceleration phase first.
v  u  at
0  15  3t
t  5 seconds
v 2  u 2  2as
02  152  2  3  s
225
6
 37.5 m
So for the acceleration phase, s  60  37.5  22.5 m .
s
s  12 (u  v)t
22.5  12 (0  15)  t
t  3 seconds
So the particle travels for a total of 5 + 3 = 8 seconds.
Example 7 : A, B and C are three points, in that order, on a straight line, with BC = 40 metres. A particle
is projected from B towards A with speed 3ms-1. The particle has a uniform acceleration of
2ms-2 towards C. Find the time taken for the particle to reach C, and the total distance
travelled by the particle on reaching C.
40m
A
X
C
B
x
By choosing the direction ABC as positive, we have u  3 ms-1 , a  2 ms-2 .
s  ut  12 at 2
40  3t  12  2  t 2
t 2  3t  40  0
(t  8)(t  5)  0
t  8, t  5
There are two solutions. We reject t  5 since negative times are not possible (what this actually means
is that with this acceleration, the particle would have been at C five seconds before arriving at B).
Therefore the time taken to reach C is 8 seconds.
The distance travelled by the particle is not the same as the displacement, because the particle changes
direction. Suppose the particle gets as far as X before it turns round, and heads back to B. At the point X,
v  0 . To calculate the displacement of X,
v 2  u 2  2as
02  (3) 2  2  2  s
0  9  4s
s  2 14 m
The distance travelled by the particle is 2  2 14  40  44 12 m .
Example 8 : A particle P, travelling at 12ms-1, passes a particle Q, travelling in the same direction at 2ms1
. At this instant Q begins accelerating at 5ms-2. Find the time taken for Q to catch up with P,
the distance travelled, and the speed at which Q is then travelling.
Suppose it takes t seconds for Q to catch up with P. In this time,
sP  12t
sQ  ut  12 at 2
 2t  12  5  t 2
 2t  2.5t 2
At this time, sP and sQ will be equal. Therefore
12t  2t  2.5t 2
10t  2.5t 2  0
2.5t (4  t )  0
t  0, t  4
There are two solutions. The solution t  0 represents the initial position. Therefore Q takes 4 seconds to
catch up with P. Substituting
sP  12  4
 48 m
vQ  u  at
 2  5 4
 22 ms -1
Q has travelled 48 metres, and is moving at 22ms-1 when it catches up with P.
Example 9 : A, B and C are three points which lie in that order on a straight road with AB = 95m
and BC = 80m. A car is travelling along the road in the direction ABC with constant
acceleration a ms−2. The car passes through A with speed u ms−1, reaches B five seconds
later, and C two seconds after that. Find the values of u and a.
We consider the phases AB and AC (not BC), since they both have the same initial velocity.
For AB,
s  ut  12 at 2
95  5u  12  a  52
95  5u  12.5a
For AC,
s  ut  12 at 2
175  7u  12  a  72
175  7u  24.5a
Solving simultaneously,
95  5u 12.5a
175  7u  24.5a
7


5


665  35u  87.5a
875  35u 122.5a
210  35a
a  6 ms -2
95  5u  12.5  6
5u  20
u  4 ms -1
p40 Ex 3A
3. Free Fall Under Gravity
When an object is dropped, it falls to Earth with uniform acceleration g. The value of g is approximately
9.8 ms-2. We often approximate this with the value 10ms-2. So if we ignore the effects of air resistance,
falling objects are subject to the uniform acceleration formulas. (In fact, this is actually a simplification at higher altitudes g is lower than at sea level, and so g is not constant for the fall of an object. However,
over short distances the change in g is very small indeed and can safely be ignored. The inaccuracy due to
ignoring air resistance is far greater, as we shall see after the first example).
Example 1 : Dissatisfied with the quality of his Chinese meal, an oriental tourist drops his plate
from a rooftop restaurant, 100m above the ground. How long does it take for the chop
suey to hit the ground, and at what speed is it then travelling?
u = 0 a = 9.8 s = 100 Find t, v.
s  ut  12 at 2
100  0  12  9.8  t 2
100
4.9
t  4.52 seconds (2 d.p.)
t2 
v 2  u 2  2as
 0  2  9.8  100
 1960
v  44.3 ms -1 (1 d.p.)
This calculation ignores air resistance, which has a significant effect on falling bodies, eventually
reducing the acceleration to zero, as the body reaches its ‘terminal velocity’. To see the importance of air
resistance, consider raindrops falling from a cloud 2km up.
u = 0 a = 9·8 s = 2000 Find v.
v 2  u 2  2as
 0  2  9.8  2000
 39200
v  198.0 ms-1 (1 d.p.)
At this speed, rain would be lethal.
Example 2 : A competitor at the highland games throws a haggis vertically into the air with a velocity of
20ms-1, letting go at a height of 2m above the ground. Find its maximum height, the time
taken to fall to the ground, and the speed on impact.
u = 20 a = −9.8 (negative since in the opposite direction to initial velocity) v = 0 Find s.
v 2  u 2  2as
0  202  2  9.8  s
400
19.6
 20.4 m (1 d.p.)
s
The maximum height above the ground is therefore 22.4 m.
On impact with the ground,
u = 20 a = −9.8 s = −2 Find t, v.
s  ut  12 at 2
2  20t  12  9.8  t 2
4.9t 2  20t  2  0
20  (20) 2  4  4.9  2
2  4.9
 4.18 seconds (2 d.p., ignoring negative root)
t
v 2  u 2  2as
 202  2  9.8  2
 439.2
v  20.9 ms -1 (1 d.p.)
Example 3 : A cannon fires a shell vertically upwards from ground level at 35ms-1. At what times is it
10m above the ground? For how long is it more than 10m high?
u = 35 a = −9.8 s = 10 Find t.
s  ut  12 at 2
10  35t  12  9.8  t 2
4.9t 2  35t  10  0
35  (35) 2  4  4.9 10
2  4.9
t  0.30 seconds, 6.84 seconds
t
The time for which the height is above 10m is 6.84 − 0.30 = 6.54 seconds.
Example 4 : A juggler throws a ball up in the air with an initial speed of 5ms-1 from a height of
1.2m. Assuming that g is 10ms-2, find the maximum height that the ball reaches above
the ground and the time it takes to reach this height.
Max height when velocity is zero
u = 5, a = -10, v = 0
Find t
v = u + at
0 = 5 – 10t
T = 0.5s
Find s
v2 = u2 + 2as
0 = 25 – 20s
s = 1.25
So max height is 1.2 + 1.25 = 2.45m
Find the time for the ball to reach the ground ig the juggler misses it!
u = 5, a = -10, s = -1.2
s = ut + ½at2
-1.2 = 5t – 5t2
-6 = 25t – 25t2
25t2 – 25t – 6 = 0
(5t + 1)(5t – 6) = 0
t = -0.2s or t = 1.2 s
Clearly we can ignore -0.2s as this is inconsistent (it actually represents the time that the ball
would have needed to be thrown at from ground level to reach the clowns hand at 5ms-1
which is when t=0). So t = 1.2s
When the first ball reaches its maximum height the juggler throws another ball with the
same speed and from the same height. Where and at what time will the balls collide/pass
each other?
Consider the displacement of the two balls separately but base both displacements on the
same origin (the position of the clowns hand)
Then the ball at the top has displacement equation: s = 0×t – 5t2 + 1.25
The new ball just being thrown has displacement equation: s = 5t – 5t2
We want to see when the displacements are the same so solving simultaneously
5t – 5t2 = – 5t2 + 1.25
5t = 1.25
t = 0.25s
at this time s = 15/16m
Example 5 : Ball A is initially 7 metres directly above ball B. Ball A is dropped from rest, and at the same
time, ball B is projected vertically upwards at 10ms-1. Find the time elapsed when, and the
position at which, the balls collide.
The two distances travelled must total 7 metres. Now
s  ut  12 at 2
s A  4.9t 2
sB  10t  4.9t 2
Therefore
4.9t 2  10t  4.9t 2  7
t  0.7 seconds
s A  4.9  0.7 2
 2.401 m
The balls collide after 0.7 seconds, 2.401 metres below the initial position of ball A.
Example 6* : A particle is dropped from a height h. It passes through points A, B and C in that order,
where AB = BC, and C is at ground level. It takes four seconds to fall from A to B, and two
seconds to fall from B to C. Taking g  10 ms-2, find AB and h.
We consider the phases AB and AC (not BC) since they have the same initial velocity.
Let AB = BC = x. For AB,
s  ut  12 at 2
x  4u  12 10  42
x  4u  80
For BC,
s  ut  12 at 2
2 x  6u  12 10  62
2 x  6u  180
x  3u  90
Comparing expressions for x,
4u  80  3u  90
u  10 ms-1
x  4 10  80
 120 m
For the phase of the motion from the point of release to A,
v 2  u 2  2as
102  02  2 10  s
100  20 s
s5 m
So AB = 120m, and h  5  2 120  245 m.
Example 7 : Two particles are projected vertically upwards from the same point at ground level. Particle
A is projected at 30ms-1, and particle B two seconds later at 40ms-1. Taking g  10 ms-2, find
when and where the particles collide. Explain how you have used the assumption that A and
B are particles in your calculations.
s A  ut  12 at 2
 30t  12  10  t 2
 30t  5t 2
Now particle B has been in motion for two seconds less that A, so since A has been in motion for t
seconds, B has been in motion for t  2 seconds.
sB  ut  12 at 2
 40(t  2)  12  10  (t  2) 2
 40t  80  5t 2  20t  20
 60t  5t 2  100
At the point of collision,
s A  sB
30t  5t 2  60t  5t 2  100
100  30t
t  3 13 seconds
Substituting,
s A  ut  12 at 2
 30  3 13  12  10  3 13
2
 44 94 m
The particles collide 3 13 seconds after the projection of A, at a height of 44 94 metres.
If A and B were not particles, but had size e.g. footballs, they would collide sooner than expected because
only their edges would have to come into contact. The particle model also enables us to ignore air
resistance and spin.