# On The Move ```Average Speed
The speed of an object is the distance it travels a certain time.
10 ms-1 means 10 m travelled in one second.
50 kmh-1 means 50 km travelled in one hour.
Speeds usually vary throughout a journey.
For this reason, we often calculate the average speed of an
object.
Measuring Average Speed
stopwatch
t
d
Mark out and measure a distance d.
Measure the time taken, to travel distance d.
Calculate the average speed using:
distance
average speed 
time
d
&divide;
v
t
Quantity
Unit
speed ( v )
metre per second ( ms-1 )
distance ( d )
metres ( m )
time ( t )
seconds ( s )
x
Example 1
A van has an average speed of 10 ms-1.
Calculate the time it takes the van to travel a distance of 125 m.
v  10 ms -1
d  125 m
t  ???
t
d
v
125

10
t  12.5 s
Results (running)
d=
m
t=
s
v = ???
d
v
t

v
ms -1
Results (walking)
d=
m
t=
s
v = ???
d
v
t

v
ms -1
Instantaneous Speed
Instantaneous speed is the speed at a particular instant.
This means it is a very short distance travelled.
Length of Card
Light Gate
Electronic
Timer
The card on the trolley cuts the light beam.
The timer starts when the beam is broken.
The timer stops when the light beam is restored.
The instantaneous speed is calculated by:
instantane ous speed 
length of card
time taken
Q1.
Why is an electronic timer used rather than a stopwatch?
Human reaction time would give an inaccurate time.
Q2.
How could a more accurate instantaneous speed be recorded?
Shorter length of card.
Example 1
A trolley with a card attached to it travels down a ramp and passes
through a light gate.
The light beam is interrupted for 0.24 s.
Calculate the instantaneous speed of the trolley as it passes through
the light gate.
card width  6 cm
 0.06 m
interrupte d time  0.24 s
instantane ous speed  ???
length of card
instantane ous speed 
time taken
0.06

0.24
instantane ous speed  0.25 ms -1
Acceleration
The acceleration of an object is its change in speed in one second.
v - u
&divide;
a
t
x
Quantity
Unit
Acceleration ( a )
Metre per second per second ( ms-2 )
Final Speed ( v )
Metre per second (ms-1 )
Initial Speed ( u )
Metre per second (ms-1 )
Time (t)
Seconds (s)
Example 1
A car accelerates from a speed of 5 ms-1 to a speed of 15 ms-1 in a
time of 2.5 seconds.
What is its acceleration?
a
u  5 ms -1
v -u
t
v  15 ms -1

t  2.5 s
a  ???
15 - 5
2.5

10
2.5
a  4 ms 2
An acceleration of 4ms-2 means that the speed increases by 4ms-1
every second.
Car
Initial Speed
(ms-1)
Final Speed
(ms-1)
Time
(s)
Acceleration
(ms-2)
Lamborghini
0
28
3.3
8.5
Chevrolet
0
28
3.7
7.6
Aston Martin
0
28
4.8
5.8
Mondeo
0
28
9.9
2.83
Deceleration
Deceleration is when an object slows down.
The formula a 
v u
is still used but u (initial speed) will be
t
greater than v (final speed).
Example 1
A car decelerates from 12 ms-1 to 2 ms-1 in 5 seconds.
Calculate the deceleration.
a
u  12 ms -1
v -u
t
v  2 ms -1

t 5s
a  ???
2 - 12
5

- 10
5
a  -2 ms 2
Example 2
A car travelling at 15 ms-1 comes to rest in a time of 4 seconds.
Calculate the car’s deceleration.
a
u  15 ms -1
v -u
t
v  0 ms -1

t4s
a  ???
0 - 15
4

- 15
4
a  -3.75 ms 2
Deceleration is always negative.
Acceleration at Credit Level
Will need to rearrange equation to calculate initial speed (u) and
final speed (v).
Final Speed (v)
v -u
a
t
v - u
&divide;
a
t
x
a t  v u
v -u  a t
v u  at
*** NEW FORMULA ***
Not on data sheet – memorise.
Example 1
A car accelerates at 2.5 ms-2 for 3 seconds.
The initial speed of the car is 4 ms-1.
Calculate the final speed of the car.
a  2.5 ms
-2
v  u  at
t 3s
 4  2.5  3
u  4 ms -1
v  ???
 4  7.5
v  11.5 ms 1
Initial Speed (u)
Use the equation
rearrange.
v u  at
and substitute the numbers and
Example 2
A car reaches a speed of 20 ms-1 after accelerating at 2 ms-2 for 4
seconds.
Calculate the initial speed of the car.
v  u  at
v  20 ms -1
a  2 ms -2
t4s
u  ???
20  u  2  4
20  u  8
u  8  20
u  20 - 8
u  12 ms -1
Speed-Time Graphs
v
v
t
acceleration
v
t
constant speed
t
deceleration
Steeper lines mean quicker acceleration / deceleration.
speed
(ms-1)
12
2
1
0
3
10
3
14
time (s)
The motion of the object is in 3-sections.
Section 1
Object accelerates from rest.
Section 2
Object travels at a constant speed of 12ms-1.
Section 3
Object decelerates and comes to a halt.
Example 1
speed
(ms-1)
20
0
11
4
time (s)
(a)
Describe the 3-parts of the journey.
(b)
Calculate the initial acceleration.
(c)
Calculate the final deceleration.
14
(a)
(b)
Describe the 3-parts of the journey.
Part 1
Object accelerates from rest to 20ms-1.
Part 2
Object travels at a constant speed.
Part 3
Object decelerates from 20ms-1 to rest.
Calculate the initial acceleration.
a
u  0 ms -1
v  20 ms
t4s
a  ???
-1
v -u
t
20 - 0

4
20

4
a  5 ms 2
(c)
Calculate the final deceleration.
v -u
a
t
v  0 ms -1
u  20 ms
t 3s
a  ???
-1
0 - 20

3

- 20
3
a  6.67 ms 2
Example 2
16
3
10
speed
(ms-1)
4
2
1
0
4
2
5
56
11
14
time (s)
(a)
Describe motion of the object for each section of the graph.
(b)
Calculate the acceleration for section 1 and 3 of the graph.
(c)
Calculate the final deceleration of the object.
(d)
What is the acceleration at section 4 of the graph?
Total Distance Travelled
The total distance travelled can be calculated from a speed-time
graph.
Total Distance
Travelled
=
Area Under Speed-Time
Graph
Example 1
speed
(ms-1)
12
2
1
0
3
10
3
time (s)
Calculate the total distance travelled.
14
Total Distance = Area Under Graph
= Area 1 + Area 2 + Area 3
Area 1
Area 2
1
bh
2
1
  3  12
2
area 
 18 m
Area 3
 7 12
 84 m
The total distance travelled is:
1
bh
2
1
  4  12
2
area 
area  l  b
18  84  24
 24 m
 126 m
The average speed over the whole journey can now be calculated.
average speed 

total distance
total time
126
14
 9 ms -1
Example 2
A driver approaches traffic lights in his car.
He sees the lights change and brakes.
The speed time graph of the car, from the instant that the lights
change is:
speed
(ms-1)
(a)
Describe the
motion of the car in
the first 0.5 s.
(b)
Calculate the
deceleration of the
car.
(c)
Calculate the
distance travelled
by the car after
the brakes were
applied.
6
0
3
0.5
time (s)
(a)
Explain the shape of the graph in the first 0.5 s.
During the first 0.5 s of the journey, the car is travelling at a
constant speed of 6 ms-1.
(b)
Calculate the deceleration of the car.
a
u  6 ms -1
t  2.5 s
v  0 ms -1
a  ???
v -u
t
0-6

2.5

-6
2.5
a  2.4 ms 2
(c)
Calculate the distance travelled by the car after the brakes
were applied.
Distance Travelled = Area Under Speed-Time Graph
distance travelled  area of triangle

1
bh
2
1
 2.5  6
2
 7.5 m

```