DISTANCE, SPEED AND ACCELERATION GRAPHS
AMD THEIR VECTOR EQUIVALENTS
We will consider motion of a particle in one dimension i.e. back and forth along a straight line.
Activity 1
Class discussion about the differences between speed and velocity, distance and displacement.
Demonstrate these differences by using the room and pacing about.
Distance and Speed are scalar quantities because they just have a size or magnitude
Displacement and Velocity are their vector equivalents that have both size and direction.
Since we are working in 1-d and can only go up /down or left/right the direction will be
indicated by a negative sign on the velocity or displacement.
Acceleration can either be a scalar or a vector there isn’t a different word for the vector quantity
of acceleration
1. Distance, Displacement and Position Graphs
Distance is how far something has travelled. It always increases (like the odometer in a car)
Displacement is the distance, measured in a straight line, of an object from its initial position.
The displacement is not necessarily the same as the distance travelled.
Activity 2
Consider the distance/time graph (Slide 2 – Lesson 1 and 2)
Give the class 10 minutes to answer the questions on Slide 3
Go thorough the answers getting the students to explain their answers.
Then show Side 3 which overlays the displacement time graph for the journey over
The distance time graph.
Discuss which of the answers to the questions are more easily obtainable and which are
less easily obtainable
The gradient of a distance/time graph tells you the speed the object is travelling at.
The gradient of a displacement/time graph tells you the velocity that the object is travelling at.
The faster the subject is moving the steeper the gradient.
(Note we can estimate the velocity or speed on a graph that curves by using a tangent to find the
gradient at a point)
Example 1 : A girl throws a ball from head height (1.5m above the ground), vertically into the
air. The ball rises to a maximum height of 10m, before landing on the ground at her
feet. Find the distance travelled by the ball and its displacement.
distance travelled  8.5  10  18.5m
displacement  1.5m
If the girl had instead rolled the ball along the ground, the distance travelled and the displacement
would have been equal.
Example 2 : A man walks four miles due East, followed by three miles due North. Find the
distance travelled and the displacement.
distance travelled  4  3  7 miles
5
displacement  42  32  5 miles
3
4
In other words, displacement is only concerned with where an object is, whereas distance is
concerned with how far it travelled getting there.
Displacement is a vector quantity - it has both magnitude and direction. For one-dimensional
motion, this means the displacement can be positive or negative.
In the diagram below, object A has a displacement of −3 units (measured from the origin O), and
object B has a displacement of 2 units. Note that distance is a scalar quantity, and therefore the
distance of A from O is simply 3 units.
A
O
B
x
Activity 3
Trolley Graph Match SWF activity
Sketch displacement/time and distance/time graphs for the
following situations:
1) The pendulum of a grand father clock that oscillates 5 cm to either side and takes 1
second to swing from one side to the other
2) The height of a cricket ball thrown vertically upwards from a height of 1m to a height of
6m if it lands on the floor after 4 seconds
3) The height of a lift as it leaves the ground floor, takes 30 seconds to reach floor 5, waits
for 1 minute and then takes 40 seconds to descend to the basement
4) The height of a bungee jumper as they fall from a bridge 50m high.
Activity 4
Get the students to describe their own situations for distance/time and displacement time graphs.
Sketch each others showing the differences in the two types of graph.
Activity 5
Show slide 10 and discuss the differences between the three graphs.
A Position/time graph is the same as a displacement/time graph except that the displacements are
not necessarily measured from the starting point of the object. Instead an origin of your choosing
is used and the objects displacement from that point is shown on the graph. (Note many people
still consider this a displacement/time graph as it shows much the same information)
Exercise 1B questions 2 to 4 pg 9
2. Speed and Velocity graphs
Velocity is the rate of change of displacement of an object, whereas speed is the rate of change of
distance. Both are measured in ms-1 (metres per second). A velocity of 5ms-1 means that an
object will travel 5 metres in every second.
Velocity is a vector quantity (and therefore in one-dimensional motion can be positive or
negative), whereas speed is a scalar quantity.
In the diagram below, A and B both have speed 2ms-1. The velocity of A is −2ms-1, since it is
travelling in the negative direction, and the velocity of B is 2ms-1, since it is travelling in the
positive direction.
2ms -1
2ms -1
A
O
B
x
The gradient of a speed/time graph tells you the acceleration (as a scalar) of the object at that
instant.
The gradient of a velocity/time graph tells you the acceleration (as a vector) of the object at that
instant The faster the subject is moving the steeper the gradient.
(Note we can estimate the acceleration on a graph that curves by using a tangent to find the
gradient at a point)
The area under a speed/time graph between two times tells you the distance travelled in that time.
The area under a velocity/time graph between two times tells you the displacement of the object
in that interval (using the start time as the origin)
Activity 6
Show slides 3 to 9 and practice interpreting and drawing speed/time graphs. The last slide
branches into velocity/time graphs as an extension (plenary)
Note that to change a velocity/time graph into a speed/time graph you just need to reflect any
negative velocities into the positive part of the graph. Note also that negative velocities mean that
the object is travelling in the opposite direction.
If the velocity is changing smoothly from negative to positive or vice-versa (ie crossing the time
axes) it means that the object is momentarily (we tend to say instantaneously) at rest and is
changing its direction of travel. At this point the displacement will have reached a localised
maximum or minimum value.
Activity 7
Interpreting Velocity/time graphs.
Give pairs a copy of all the graphs in this power point and a list of the questions
Example 1 : A cyclist leaves home O and rides along a straight road with a constant acceleration.
After 10 seconds, he has reached a point A with a speed of 15ms-1, and he maintains
45
40
35
30
25
20
15
10
5
5
10
15
this speed for a further 20 seconds, until he reaches B, before retarding uniformly to
5
10
15
20
25
30
35
40
45
rest at C. The whole journey takes 45 seconds. Sketch a speed-time graph for the
journey and find
a) his acceleration for the journey from O to A.
b) his retardation for the journey from B to C.
c) the total distance travelled from O to C.
v
A
B
15
10
5
C
O
5
10
15
20
25
30
35
a) The acceleration is the gradient of the graph from O to A.
15
10
 1.5 ms -2
b) The acceleration is the gradient of the graph from B to C.
acceleration 
15
15
 1 ms-2
acceleration 
40
45
t
So the retardation is 1ms-2.
c) The total distance travelled is the area under the graph.
distance  12 (20  45) 15
 487.5 m
Example 2 : A car travelling along a straight road passes point A when t = 0, and maintains a
constant speed until t = 24 seconds. The driver then applies the brakes and the car
retards uniformly to rest at point B. Before the brakes were applied the car had
travelled 54 of the total distance AB. Sketch the speed-time graph and calculate the
time taken for the car to travel from A to B.
v
u
A
P
B
O
24
T
t
Let the initial speed be u, and the time taken for the journey be T.
distance travelled at constant speed  24u
total distance travelled  12 (24  T )u
So we have
24u  54  12 (24  T )u
24  52 (24  T )
60  24  T
T  36 seconds
The time taken is 36 seconds.
There is a much quicker, geometrical solution to this problem.
v
x
4x
x
O
T
24
t
The dotted rectangle has half the area of the rectangle to the left, and therefore half the width. Its
width must therefore be 12, and so T is 36.
Example 3 : A man is jogging along a straight road at a constant speed of 4ms-1. He passes a
friend with a bicycle who is standing at the side of the road and 20 seconds later the
friend starts cycling to catch him up. The cyclist accelerates at a uniform rate of
3ms-2 until he reaches a speed of 12ms-1. He then maintains a constant speed.
a) On the same diagram, sketch the speed-time graphs for the jogger and cyclist.
b) Find the time that elapses before the cyclist reaches the jogger.
a) Let the cyclist reach the jogger at time T.
v
12
cyclist
4
jogger
20
24
T
t
b) We require the distance travelled by the jogger to be equal to the distance travelled by the
cyclist at time T. So the areas under the two graphs up to time T must be equal.
4T  12   (T  24)  (T  20)   12
4T  6(2T  44)
4T  12T  264
264  8T
T  33 seconds
Example 4 : Two stations A and B are a distance of 6x metres apart along a straight track. A
train starts from rest at A and accelerates uniformly to a speed v ms-1, covering a
distance x metres. The train then maintains this speed until it has travelled a further
3x metres. It then retards uniformly to rest at B. Make a sketch of the velocity-time
graph for the motion and show that if T is the time taken for the train to travel from
9x
A to B, then T 
seconds.
v
v
3x
v
v
x
3x
2x
B
A
T
The time taken for the steady speed phase is clearly
t
3x
.
v
Using the formula for the area of a trapezium,
area  12 (a  b)h
 3x

6 x  12   T  v
 v

12 x  3 x  Tv
9 x  Tv
T
9x
v
Alternatively, we can solve the problem
v
geometrically.
v
Comparing expressions for the area of the
rectangle,
Tv  9 x
T
9x
v
x
2x
x
A
3x
2x
B
T
t
Example 5 : Now solve example 4 from section 8 the easy way!
v
15
60
T t
area of triangle  12 bh
60  12  T 15
120  15T
T  8 seconds
Notice how much easier this is, and that the retardation in the second phase is not even needed!
Exercise 1D questions 1 to 8 pg 16
3. Acceleration/time graphs
Acceleration is the rate of change of velocity of an object. It is measured in ms-2 (metres per
second squared). An acceleration of 5ms-2 means that the velocity of an object will increase by
5ms-1 in every second. Similarly, an acceleration of −5ms-2 means that the velocity of an object
will decrease by 5ms-1 in every second. An acceleration of −5ms-2 is often known as a retardation
of 5ms-2.
Example 1 : An object is travelling with a velocity of 3ms-1. It is accelerating at 2.5ms-2. Find its
velocity after 7 seconds.
velocity  3  (7  2.5)
 20.5 ms -1
Example 2 : An object is travelling with a velocity of 2ms-1. It is accelerating at −1.5ms-2. Find
its velocity after 4 seconds.
velocity  2  (4  1.5)
 4 ms -1
Activity 8
Instead of using the examples below use the excel spreadsheet
displacement+velocity+acceleration locator to test pupils understanding of the vector quantities
for acceleration, velocity and displacement.
Example 3 : Draw a diagram to illustrate an object for which...
s : negative
v : positive
a : positive
speeding up
x
O
Example 4 : Draw a diagram to illustrate an object for which...
s : positive
v : positive
a : negative
O
x
slowing down
Example 5 : Draw a diagram to illustrate an object for which...
s : positive
v : negative
a : negative
O
x
speeding up
It may seem strange that an object with negative acceleration is actually speeding up. This
happens because the object is moving with negative velocity, and decreasing a negative velocity
actually increases its magnitude (i.e. its speed). For example, if the velocity is −3ms-1 and the
acceleration is −1ms-2, then one second later the velocity will be −4ms-2, which is faster than
−3ms-2.
Example 6 : Draw a diagram to illustrate an object for which...
s : negative
v : positive
a : zero
constant speed
O
x
Example 7 : Draw a diagram to illustrate an object for which...
s : negative
v : zero
a : positive
O
not moving, about to move to the right
x
Example 8 : Draw a diagram to illustrate an object for which...
s : zero
v : negative
a : positive
slowing down O
x
The area under an acceleration/time graph between two times tells you the increase/decrease in
velocity between those times. Note that when acceleration is zero velocity is at a localised
maximum or minimum value.
Exercise 1C questions 1 to 6 pg 11
140
40
30
20
23410
10
20
30
4321
4. Uniform Velocity
A particle travels with uniform velocity if its speed and direction are constant. The graph of
displacement against time is a straight line, and the gradient is the velocity.
s
40
velocity 
30
20
10
 5 ms -1
2
10
2
10
1
2
3
4
t
5. Average Velocity
The velocity of an object may change over time. We define
displacement
average velocity 
time taken
distance
average speed 
time taken
Example 1 : A car travels for 4 hours at 100km/h in one direction, and then for 160km at 80km/h
in the opposite direction. Find the average velocity and the average speed of the car.
In the first stage of the motion, the car travels 400km.
160
 2 hours.
The second stage of the motion takes
80
displacement
time taken
400  160

6
 40 km/h
average velocity 
distance
time taken
400  160

6
1
 93 3 km/h
average speed 
We can interpret these figures thus. If the car is to get to its
final destination by travelling in a straight line, it should move
at a constant 40 km/h. If the car is to get to its final
destination by its actual route, it should move at a constant
93 13 km/h.
1
40
30
20
10
4
3
2
10
20
30
40
1
2
3
4
In the displacement-time graph opposite, the average velocity
is the gradient of the dotted line. It is not the average of the
individual velocities.
s
400
300
200
100
1
2
3
4
5
6
t
6. Uniform Acceleration
An object travels with uniform acceleration if its acceleration is constant. The graph of velocity
against time is a straight line and the gradient is the acceleration.
v
40
acceleration 
30
20
10
2
10
1
2
3
4
t
10
 5 ms -2
2