Teach A Level Teach A Level MathsMaths

Displacement Displacement andandVelocityVelocity

Volume 4: Mechanics 1Volume 4: Mechanics 1Displacement and Displacement and

VelocityVelocity

and then 2 m left.The distance I have moved is 4 m BUT I’m back where I started.

My displacement is 0 m

Displacement measures the change in position and is a vector.

If I move 2 more metres to the left, my displacement from the start is 2 m ( although the distance I’ve moved is now 6 m ).

xStart

Suppose I move 2 m to the right in a straight line . . .

Using the example from the previous presentation, instead of graphs showing distance and speed against time, we can draw displacement and velocity graphs.

Speed is the rate of change of distance and velocity is the rate of change of

displacement.

Displacement and velocity are both vector quantities so have direction as well as magnitude.

When we talk about “rate of change” for distance and displacement we mean how the quantities change with time.

Speed is the magnitude of velocity.

time (s)

displacement (m)

time (s)

velocity (ms-1)

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

time (s)

displacement (m)

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

xstart

time (s)

displacement (m)

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

xstart

time (s)

displacement (m)

Here I’m back at the start.

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

xstart

time (s)

displacement (m)

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

xstart

time (s)

displacement (m)

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2The negative displacement shows where I am on the left, rather than the right, of the start.

xstart

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

time (s)

velocity (ms-1)

xstart

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

time (s)

velocity (ms-1)

xstart

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

time (s)

velocity (ms-1)

xstart

The negative velocity shows movement in the opposite direction ( towards the left of the start instead of towards the right ).

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

time (s)

velocity (ms-1)

xstart

MoveSpee

d(m s-1)

Time(s)

2 m right

1 2

at rest 0 22 m left 1 2at rest 0 2

2 m left 1 2

time (s)

velocity (ms-1)

xstart

I’m still walking to the left.

time (s)

velocity (ms-1)

time (s)

displacement (m)

On a displacement-time graph gradient gives velocity.

time (s)

displacement (m)

On the displacement graph, the two sections with negative gradients give the velocity as 1.

22

Tip: When the gradient is negative, as the quantity on the time axis increases . . . the quantity on the displacement axis

decreases.A line with negative gradient falls

backwards

time (s)

displacement (m)

Average velocity

total displacementtotal

timee.g. Using the graph, find the average velocity for (a) the entire journey, and(b) the 1st 6 seconds of the motion Solution:(a) Average velocity2

10 0·2 m s

-

1

(b) Average velocity for 1st 6 seconds

0 6

0

Average velocity is zero when the displacement is

zero.

time (s)

velocity (ms-1)

Decide how the parts below the axis give us displacement.

Ans: We find the area . . .

We can find displacement from a velocity-time graph but we have to be careful.

e.g.

but since area is positive and displacement is negative we must insert a minus sign.

time (s)

velocity (ms-1)

1

Tip: For areas below the time axis the negative velocity reminds us that the displacement is

also negative.

time (s)

velocity (ms-1)

Area A 2 1 2

The total displacement Area A – Area B – Area C

The total displacement 2 2 2 2

m

Area B Area C Area A 2

A

B C

We can also find the distance travelled from the velocity/time graph.

time (s)

velocity (ms-1)

To find the total distance, we just ignore the negative signs on the displacements where the graph lies below the axis.

The total distance travelled 2 + 2 + 2 6 m

The total displacement 2 2 2 2

m

SUMMARY

Displacement measures the change in position and is a vector.

Velocity is the rate of change of displacement and is also a vector.

Motion in a straight line can be illustrated on graphs.

continued

time (s)

displacement (m) negative

velocity

time (s)

velocity (ms-1)

negative displacement

6

6

Areas below the axis give negative displacements. If negative signs for displacement are ignored the area gives distance.

On a velocity-time graph we can find displacement and distance from area.

On a displacement-time graph gradient gives velocity.

SUMMARY

(i) Constant speed of 4 m s-1 for 4 s from A to C.(ii) Constant speed of 2 m s-1 for 2 s from C to B.(iii) At rest at B for 2 s(b)Use the graph to find the displacement

from A after 6 s.(c) Complete the graph if the particle returns

to A in a further 2 s.

(a) Sketch a velocity-time graph to show the following motion of a particle moving for 10 s in a straight line ABC where B lies between A and C.

Your sketch need not use squared paper nor be to scale but you must mark the key values on the axes.

EXERCISE

Since B is between A and C the particle changes direction so the velocity is negative.

(a)(i) Constant speed of 4 m s-1 for 4 s from A to C.(ii) Constant speed of 2 m s-1 for 2 s from C to

B.(iii) At rest at B for 2 s.

Solution:

velocity (ms

-1)

time (s)

4

4

862

EXERCISE

velocity (ms

-1)

time (s)

4

4

Solution:

862

(b)Find the displacement from A after 6 s.

16

4

The displacement after 6 s 16 – 4 12 m

A to C

C to B

EXERCISE

velocity (ms

-1)

time (s)

4

4

Solution:

862

16

4

A to C

C to B

(c) Complete the graph if the particle returns to A in a further 2 s.

10

6

The displacement of B from ASo, the particle must move at 6 m s

-1 to return to A in 2 s.

12 m

EXERCISE

velocity (ms

-1)

time (s)

4

4

Solution:

862

16

4

A to C

C to B

(c) Complete the graph if the particle returns to A in a further 2 s.

10

6

We can check this result by noticing that the total displacement is zero.

12

B to A

EXERCISE

The slides that follow are in a form suitable for photocopying for example 1 and the summary.

212 m left

20at rest

212 m left

20at rest

2 m right

Move

21

Time(s)

Speed(m s -1)

Displacement and Velocity - Data and Graphs for e.g.1

time (s)

displacement (m)

time (s)

velocity (ms-1)

Summary

DISPLACEMENT AND VELOCITY

TEACH A LEVEL MATHS – MECHANICS 1

Displacement measures the change in position and is a vector. Velocity is the rate of change of displacement and is also a vector. Motion in a straight line can be illustrated on graphs.

time (s)

displacement (m)negative velocity

time (s)

velocity (ms-1)

negative displacement

6

6

Areas below the axis give negative displacements. If negative signs for displacement are ignored the area gives distance.

On a velocity-time graph we can find displacement and distance from area.

On a displacement-time graph gradient gives velocity.