21ms-1
Introduction
• This chapter you will learn the SUVAT
equations
• These are the foundations of many of
the Mechanics topics
• You will see how to use them to use
many types of problem involving motion
Kinematics of a Particle moving in a
Straight Line
You will begin by learning two of the
SUVAT equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑎=
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
𝑣−𝑢
𝑡
Multiply by t
Replace with the
appropriate letters.
 Change in velocity =
final velocity – initial
velocity
𝑎𝑡 = 𝑣 − 𝑢
Add u
𝑎𝑡 + 𝑢 = 𝑣
𝑣 = 𝑢 + 𝑎𝑡
This is the
usual form!
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 × 𝑡𝑖𝑚𝑒
𝑠=
𝑢+𝑣
𝑡
2
Replace
with the
appropriate
letters
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning two of the
SUVAT equations
You need to consider using negative numbers in
some cases
Positive direction
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
𝑢+𝑣
𝑡
2
The particle is to the left of
the point O, which is the
negative direction
The particle is moving at
2.5ms-1 in the positive direction
2.5ms-1
6ms-1
P
Q
4m
O
3m
If we are measuring displacements from O, and left to right
is the positive direction…
For particle P:
𝑠 = −4𝑚
𝑣 = 2.5𝑚𝑠 −1
The particle is to
the right of the
point O, which is
the positive
direction
The particle is moving at 6ms-1
in the negative direction
For particle Q:
𝑠 = 3𝑚
𝑣 = −6𝑚𝑠 −1
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
𝑣 = 𝑢 + 𝑎𝑡
𝑢+𝑣
𝑠=
𝑡
2
A particle is moving in a straight line from A to B with constant
acceleration 3ms-2. Its speed at A is 2ms-1 and it takes 8 seconds to
move from A to B. Find:
a) The speed of the particle at B
b) The distance from A to B
2ms-1
Start with a
diagram
A
𝑠 =?
𝑢=2
B
𝑣 =?
𝑣 = 𝑢 + 𝑎𝑡
𝑣 = 2 + (3 × 8)
𝑣 = 26𝑚𝑠 −1
𝑎=3
Fill in the
values you
know
𝑡=8
Write out ‘suvat’ and
fill in what you know
For part a) we need
to calculate v, and we
know u, a and t…
Remember to
include units!
You always need to set up the question in this
way. It makes it much easier to figure out what
equation you need to use (there will be more to
learn than just these two!)
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
𝑢+𝑣
𝑡
2
A particle is moving in a straight line from A to B with constant
acceleration 3ms-2. Its speed at A is 2ms-1 and it takes 8 seconds to
move from A to B. Find:
a) The speed of the particle at B – 26ms-1
b) The distance from A to B
2ms-1
A
𝑠 =?
𝑢=2
B
𝑣 =?
= 26
𝑢+𝑣
𝑠=
𝑡
2
𝑠=
2 + 26
×8
2
𝑠 = 14 × 8
𝑠 = 112𝑚
𝑎=3
Fill in the
values you
know
𝑡=8
For part b) we need
to calculate s, and we
know u, v and t…
Show
calculations
Remember
the units!
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
A cyclist is travelling along a straight road. She accelerates at a
constant rate from a speed of 4ms-1 to a speed of 7.5ms-1 in 40
seconds. Find:
a) The distance travelled over this 40 seconds
b) The acceleration over the 40 seconds
4ms-1
𝑠 =?
𝑢=4
𝑠=
𝑢+𝑣
𝑡
2
𝑠=
4 + 7.5
× 40
2
𝑣 = 𝑢 + 𝑎𝑡
𝑢+𝑣
𝑠=
𝑡
2
7.5ms-1
𝑣 = 7.5
𝑎 =?
𝑡 = 40
Sub in the
values you
know
Draw a diagram
(model the cyclist as
a particle)
Write out ‘suvat’ and
fill in what you know
We are calculating s,
and we already know
u, v and t…
Remember
units!
𝑠 = 230𝑚
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
A cyclist is travelling along a straight road. She accelerates at a
constant rate from a speed of 4ms-1 to a speed of 7.5ms-1 in 40
seconds. Find:
a) The distance travelled over this 40 seconds – 230m
b) The acceleration over the 40 seconds
4ms-1
𝑠=
𝑠 =?
230
𝑢=4
7.5ms-1
𝑣 = 7.5
𝑣 = 𝑢 + 𝑎𝑡
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
𝑢+𝑣
𝑡
2
7.5 = 4 + 40𝑎
7.5 = 40𝑎
𝑎 = 0.0875𝑚𝑠 −2
𝑎 =?
Sub in the
values you
know
𝑡 = 40
Draw a diagram
(model the cyclist as
a particle)
Write out ‘suvat’ and
fill in what you know
For part b, we are
calculating a, and we
already know u, v and
t…
Subtract 4
Divide by
40
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
𝑢+𝑣
𝑡
2
A particle moves in a straight line from a point A to B with constant
deceleration of 1.5ms-2. The speed of the particle at A is 8ms-1 and the
speed of the particle at B is 2ms-1. Find:
a) The time taken for the particle to get from A to B
b) The distance from A to B
8ms-1
2ms-1
Draw a diagram
A
𝑠 =?
𝑢=8
𝑣 = 𝑢 + 𝑎𝑡
2 = 8 − 1.5𝑡
−6 = −1.5𝑡
4=𝑡
B
𝑣=2
𝑎 = −1.5
Sub in the
values you know
𝑡 =?
Write out ‘suvat’ and
fill in what you know
As the particle is
decelerating, ‘a’ is
negative
Subtract 8
Divide by -1.5
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
𝑢+𝑣
𝑡
2
A particle moves in a straight line from a point A to B with constant
deceleration of 1.5ms-2. The speed of the particle at A is 8ms-1 and the
speed of the particle at B is 2ms-1. Find:
a) The time taken for the particle to get from A to B – 4 seconds
b) The distance from A to B
8ms-1
2ms-1
Draw a diagram
A
𝑠 =?
𝑢=8
𝑢+𝑣
𝑠=
𝑡
2
𝑠=
8+2
×4
2
B
𝑣=2
𝑎 = −1.5
Sub in the
values you know
𝑡=
=?4
Write out ‘suvat’ and
fill in what you know
As the particle is
decelerating, ‘a’ is
negative
Calculate the
answer!
𝑠 = 20𝑚
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
After reaching B the particle continues to move along the straight line
with the same deceleration. The particle is at point C, 6 seconds after
passing through A. Find:
a) The velocity of the particle at C
b) The distance from A to C
8ms-1
2ms-1
A
𝑠 =?
B
𝑢=8
𝑣 =?
𝑣 = 𝑢 + 𝑎𝑡
𝑣 = 𝑢 + 𝑎𝑡
𝑢+𝑣
𝑠=
𝑡
2
?
𝑣 = 8 − (1.5 × 6)
𝑣 = −1𝑚𝑠 −1
𝑎 = −1.5
Update the
diagram
C
𝑡=6
Write out
‘suvat’ using
points A and C
Sub in the
values
Work it
out!
As the velocity is negative, this means the
particle has now changed direction and is
heading back towards A! (velocity has a
direction as well as a magnitude!)
The velocity is 1ms-1 in the direction C to A…
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
After reaching B the particle continues to move along the straight line
with the same deceleration. The particle is at point C, 6 seconds after
passing through A. Find:
a) The velocity of the particle at C - -1ms-1
b) The distance from A to C
8ms-1
A
𝑠 =?
𝑠=
𝑣 = 𝑢 + 𝑎𝑡
𝑢+𝑣
𝑠=
𝑡
2
2ms-1
𝑠=
𝑢=8
?
B
𝑣𝑣==?
−1
𝑢+𝑣
𝑡
2
8−1
×6
2
𝑎 = −1.5
Update the
diagram
C
𝑡=6
Write out
‘suvat’ using
points A and C
Sub in the
values
Work it
out!
𝑠 = 21𝑚
It is important to note that 21m is the distance from A to C
only…
 The particle was further away before it changed
direction, and has in total travelled further than 21m…
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
A car moves from traffic lights along a straight road with constant
acceleration. The car starts from rest at the traffic lights and 30
seconds later passes a speed trap where it is travelling at 45 kmh-1. Find:
a) The acceleration of the car
b) The distance between the traffic lights and the speed-trap.
0ms-1
45kmh-1
Draw a diagram
Lights
Trap
Standard units to use are metres and seconds, or kilometres and hours
 In this case, the time is in seconds and the speed is in kilometres
per hour
 We need to change the speed into metres per second first!
𝑣 = 𝑢 + 𝑎𝑡
𝑢+𝑣
𝑠=
𝑡
2
45𝑘𝑚ℎ−1
Multiply by 1000 (km to m)
45,000𝑚ℎ−1
12.5𝑚𝑠 −1
Divide by 3600 (hours to seconds)
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
A car moves from traffic lights along a straight road with constant
acceleration. The car starts from rest at the traffic lights and 30
seconds later passes a speed trap where it is travelling at 45 kmh-1. Find:
a) The acceleration of the car
b) The distance between the traffic lights and the speed-trap.
0ms-1
45kmh-1 = 12.5ms-1
Draw a diagram
Lights
𝑠 =?
𝑢=0
Trap
𝑣 = 12.5
𝑣 = 𝑢 + 𝑎𝑡
𝑣 = 𝑢 + 𝑎𝑡
𝑢+𝑣
𝑠=
𝑡
2
12.5 = 0 + 30𝑎
5
𝑚𝑠 −2 = 𝑎
12
𝑎 =?
𝑡 = 30
Write out ‘suvat’ and
fill in what you know
Sub in the
values
Divide by
30
You can use
exact answers!
2A
Kinematics of a Particle moving in a
Straight Line
You will begin by learning
two of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocity
v = Final velocity
a = Acceleration
t = Time
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
𝑢+𝑣
𝑡
2
A car moves from traffic lights along a straight road with constant
acceleration. The car starts from rest at the traffic lights and 30
seconds later passes a speed trap where it is travelling at 45 kmh-1. Find:
a) The acceleration of the car
b) The distance between the traffic lights and the speed-trap.
0ms-1
45kmh-1 = 12.5ms-1
Draw a diagram
Lights
𝑠 =?
𝑠=
𝑠=
𝑢=0
Trap
𝑣 = 12.5
𝑢+𝑣
𝑡
2
0 + 12.5
× 30
2
5
𝑎=
=?12 𝑡 = 30
Write out ‘suvat’ and
fill in what you know
Sub in
values
Work it
out!
𝑠 = 187.5𝑚
2A
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae linking
different combination of ‘SUVAT’, for a
particle moving in a straight line with
constant acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 − 𝑢 = 𝑎𝑡
𝑢+𝑣
𝑡
2
2
2
Divide by a
𝑣−𝑢
=𝑡
𝑎
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
Subtract u
𝑠=
𝑣 = 𝑢 + 2𝑎𝑠
𝑠=
𝑢+𝑣
𝑡
2
𝑢+𝑣
2
𝑣−𝑢
𝑎
𝑣 2 − 𝑢2
𝑠=
2𝑎
2𝑎𝑠 = 𝑣 2 − 𝑢2
2
𝑢 + 2𝑎𝑠 = 𝑣
Replace t with the
expression above
2
𝑣 2 = 𝑢2 + 2𝑎𝑠
Multiply numerators and
denominators
Multiply by 2a
Add u2
This is the way it is
usually written!
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae linking
different combination of ‘SUVAT’, for a
particle moving in a straight line with
constant acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑢+𝑣
𝑠=
𝑡
2
𝑣 2 = 𝑢2 + 2𝑎𝑠
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
𝑠=
𝑢+𝑣
𝑡
2
𝑢 + 𝑢 + 𝑎𝑡
𝑠=
𝑡
2
𝑠=
2𝑢 + 𝑎𝑡
𝑡
2
1
𝑠 = 𝑢 + 𝑎𝑡 𝑡
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
Replace ‘v’ with ‘u + at’
Group terms on the
numerator
Divide the numerator
by 2
Multiply out the
bracket
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae linking
different combination of ‘SUVAT’, for a
particle moving in a straight line with
constant acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑠=
𝑢+𝑣
𝑡
2
𝑣 2 = 𝑢2 + 2𝑎𝑠
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
𝑣 = 𝑢 + 𝑎𝑡
𝑣 − 𝑎𝑡 = 𝑢
Subtract ‘at’
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
𝑠 = (𝑣 − 𝑎𝑡)𝑡 + 𝑎𝑡 2
2
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2 + 𝑎𝑡 2
2
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
Replace ‘u’ with ‘v - at’
from above’
Multiply out the
bracket
Group up the at2
terms
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae
linking different combination of
‘SUVAT’, for a particle moving in
a straight line with constant
acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 2 = 𝑢2 + 2𝑎𝑠
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
𝑠=
A particle is moving in a straight line from A to B with constant
acceleration 5ms-2. The velocity of the particle at A is 3ms-1 in the
direction AB. The velocity at B is 18ms-1 in the same direction. Find the
distance from A to B.
3ms-1
𝑢+𝑣
𝑡
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
18ms-1
Draw a diagram
A
𝑠 =?
B
𝑢=3
𝑣 = 18
𝑣 2 = 𝑢2 + 2𝑎𝑠
2
2
18 = 3 + 2(5)𝑠
324 = 9 + 10𝑠
315 = 10𝑠
31.5𝑚 = 𝑠
𝑎=5
𝑡 =?
Replace v, u and a
Write out ‘suvat’
with the
information given
We are
calculating s,
using v, u and a
Work out terms
Subtract 9
Divide by 10
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae
linking different combination of
‘SUVAT’, for a particle moving in
a straight line with constant
acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 2 = 𝑢2 + 2𝑎𝑠
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
𝑠=
A car is travelling along a straight horizontal road with a constant
acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it passes a
pillar box. 12 seconds later the car passes a lamp post. Find:
a) The distance between the pillar box and the lamp post
b) The speed with which the car passes the lamp post
8ms-1
𝑢+𝑣
𝑡
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
Pillar
Box
𝑠 =?
𝑢=8
Draw a diagram
Lamp
Post
𝑣 =? 𝑎 = 0.75 𝑡 = 12
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
𝑠 = (8 × 12) + (0.75 × 122 )
2
Write out ‘suvat’
with the
information given
We are
calculating s,
using u, a and t
Replace u, a
and t
Calculate
𝑠 = 150𝑚
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae
linking different combination of
‘SUVAT’, for a particle moving in
a straight line with constant
acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 2 = 𝑢2 + 2𝑎𝑠
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
𝑠=
A car is travelling along a straight horizontal road with a constant
acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it passes a
pillar box. 12 seconds later the car passes a lamp post. Find:
a) The distance between the pillar box and the lamp post – 150m
b) The speed with which the car passes the lamp post
8ms-1
𝑢+𝑣
𝑡
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
Draw a diagram
Pillar
Box
𝑠 =?
𝑢=8
Lamp
Post
𝑣 =? 𝑎 = 0.75 𝑡 = 12
𝑣 = 𝑢 + 𝑎𝑡
𝑣 = 8 + (0.75 × 12)
Write out ‘suvat’
with the
information given
We are
calculating v,
using u, a and t
Replace u, a
and t
Calculate
𝑣 = 17𝑚𝑠 −1
Often you can use an answer you have calculated later
on in the same question. However, you must take care
to use exact values and not rounded answers!
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae
linking different combination of
‘SUVAT’, for a particle moving in
a straight line with constant
acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 2 = 𝑢2 + 2𝑎𝑠
𝑠=
A particle is moving in a straight horizontal line with constant
deceleration 4ms-2. At time t = 0 the particle passes through a point O
with speed 13ms-1, travelling to a point A where OA = 20m. Find:
a) The times when the particle passes through A
b) The total time the particle is beyond A
c) The time taken for the particle to return to O
𝑢+𝑣
𝑡
2
Draw a diagram
13ms-1
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
O
A
𝑠 = 20 𝑢 = 13 𝑣 =? 𝑎 = −4
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
20 = (13)𝑡 + (−4)𝑡 2
2
We have 2 answers. As the
acceleration is negative, the
particle passes through A, then
changes direction and passes
through it again!
20 = 13𝑡 − 2𝑡 2
2
2𝑡 − 13𝑡 + 20 = 0
(2𝑡 − 5)(𝑡 − 4) = 0
𝑡 = 2.5 𝑜𝑟 4
𝑡 =?
Replace s, u
and a
Write out ‘suvat’
with the
information given
We are
calculating t,
using s, u and a
Simplify terms
Rearrange and set equal to 0
Factorise (or use the quadratic formula…)
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae
linking different combination of
‘SUVAT’, for a particle moving in
a straight line with constant
acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 2 = 𝑢2 + 2𝑎𝑠
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
𝑠=
𝑢+𝑣
𝑡
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
A particle is moving in a straight horizontal line with constant
deceleration 4ms-2. At time t = 0 the particle passes through a point O
with speed 13ms-1, travelling to a point A where OA = 20m. Find:
a) The times when the particle passes through A – 2.5 and 4 seconds
b) The total time the particle is beyond A
c) The time taken for the particle to return to O
Draw a diagram
13ms-1
O
𝑠 = 20 𝑢 = 13 𝑣 =? 𝑎 = −4
A
𝑡 =?
The particle passes through A at 2.5
seconds and 4 seconds, so it was
beyond A for 1.5 seconds…
Write out ‘suvat’
with the
information given
We are
calculating t,
using s, u and a
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae
linking different combination of
‘SUVAT’, for a particle moving in
a straight line with constant
acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 2 = 𝑢2 + 2𝑎𝑠
𝑠=
A particle is moving in a straight horizontal line with constant
deceleration 4ms-2. At time t = 0 the particle passes through a point O
with speed 13ms-1, travelling to a point A where OA = 20m. Find:
a) The times when the particle passes through A – 2.5 and 4 seconds
b) The total time the particle is beyond A – 1.5 seconds
c) The time taken for the particle to return to O
𝑢+𝑣
𝑡
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
Draw a diagram
13ms-1
O
A
𝑠𝑠==20
0 𝑢 = 13 𝑣 =? 𝑎 = −4
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
0 = (13)𝑡 + (−2)𝑡
0 = 13𝑡 − 2𝑡 2
The particle is at O when t = 0
seconds (to begin with) and is
at O again when t = 6.5 seconds
2𝑡 2 − 13𝑡 = 0
𝑡(2𝑡 − 13) = 0
𝑡 = 0 𝑜𝑟 6.5
2
𝑡 =?
Write out ‘suvat’
with the
information given
The particle
returns to O
when s = 0
Replace s, u and a
Simplify
Rearrange
Factorise
2B
Kinematics of a Particle moving in a
Straight Line
You can also use 3 more formulae
linking different combination of
‘SUVAT’, for a particle moving in
a straight line with constant
acceleration
𝑣 = 𝑢 + 𝑎𝑡
𝑣 2 = 𝑢2 + 2𝑎𝑠
1
𝑠 = 𝑣𝑡 − 𝑎𝑡 2
2
𝑠=
A particle is travelling along the x-axis with constant deceleration
2.5ms-2. At time t = O, the particle passes through the origin, moving
in the positive direction with speed 15ms-1. Calculate the distance
travelled by the particle by the time it returns to the origin.
15ms-1
Draw a diagram
𝑢+𝑣
𝑡
2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
O
𝑠 =?
X
𝑢 = 15 𝑣 = 0 𝑎 = −2.5 𝑡 =?
The total distance
travelled will be double the
distance the particle
reaches from O (point X)
 At X, the velocity is 0
𝑣 2 = 𝑢2 + 2𝑎𝑠
02 = 152 + 2(−2.5)𝑠
Replace v,
u and a
Simplify
We are
calculating s,
using u, v and a
0 = 225 − 5𝑠
5𝑠 = 225
𝑠 = 45𝑚
𝑇𝑜𝑡𝑎𝑙 𝑠 = 90𝑚
Add 5s
Divide by 5
45m is the distance from O
to X. Double it for the total
distance travelled
2B