©John Parkinson

1

©John Parkinson

2

Distance travelled - s

Time taken - tVelocity - v

v =s

t v

s / t

Velocity = Speed in a Specified Direction

Constant Velocity

©John Parkinson

3

N

100 m

in 4 seconds

Distance travelled = ? 100 m

Displacement = ? 100 m to the East

Speed = ? Speed = 100/4 = 25 m s-1

Velocity = ? Velocity = 25 m s-1 to the East

©John Parkinson

4

DISPLACEMENT – TIME GRAPHS

Constant velocity

Displacement - s

Time - t

What will the graph look like?

GRADIENT = ?

Δt

Δs

t

sv

VELOCITY

©John Parkinson

5

Displacement - s

Time - t

What about this graph?

A body at rest

Displacement - s

Time - t

And this graph?

The gradient is …….?increasing

Δs

Δt

The body must be ……..?accelerating

©John Parkinson

6

1 3

2

A

VELOCITY – TIME GRAPHS

Velocity - v

Time - t

Velocity - v

Time - t

This body has a constant or uniform ………?acceleration

Δv

Δt

The gradient = ?the acceleration

t

va

1 = …… ?Uniform acceleration

2 = …… ?Constant velocity

3 = …… ?Uniform retardation [deceleration]

Area under the graph = A

= …….. ?DISTANCE TRAVELLED

©John Parkinson

7

Velocity – v/ms-1

Time – t/s

30

20 50 80

QUESTION The graph represents the motion of a tube train between two stations

Find

1. The acceleration

2. The maximum velocity

3. The retardation

4. The distance travelled

1. The acceleration = the initial gradient = 30÷20 = 1.5 m s-2

2. The maximum velocity is read from the graph = 30 m s-1

3. The retardation = the final gradient = -30 ÷ [80-50] = -1.0 m s-2

4. The distance travelled = the area under the graph

=½ x 20 x 30 + 30 x 30 + ½ x 30 x 30 = 1650 m

©John Parkinson

8

What will the distance – time, velocity - time and acceleration time graphs look like for this bouncing ball?

s1

s2

Displacement - s

Time - t

Velocity - v

Time - t

s1

s2

©John Parkinson

9

Acceleration - a

Time - t

Velocity - v

Time - t

9.81ms-2

©John Parkinson

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Velocity

Time

What might this graph represent?

Can you draw an acceleration time graph for this motion?

Terminal Velocity

Acceleration

Time

9.81 m s-1

©John Parkinson

11

EQUATIONS OF MOTIONEQUATIONS OF UNIFORM ACCELERATION

For Constant Velocity

DISTANCE = VELOCITY x TIME

VELOCITY =DISTANCE

TIME

t

sv

If Velocity is not constant , this equation just gives the average velocity for the journey

tvs

©John Parkinson

12

EQUATIONS OF MOTIONEQUATIONS OF MOTION

For UNIFORM ACCELERATION

SYMBOLS

• a = ACCELERATION

• u = INITIAL VELOCITY

• v = FINAL VELOCITY

• s = DISTANCE TRAVELLED

• t = TIME TAKEN

©John Parkinson

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For UNIFORM ACCELERATION

1. Distance travelled = average velocity times the time taken

s = u + v

2t

Velocity

Time

u

v

t

2. Acceleration = the change in velocity per second

a =v - u

t

Rearranging

v = u + at

©John Parkinson

14

3. Substituting equation [2] into equation [1]

s = u + v2

tv = u + at

2

2

2

2atutt

atuus

s = ut + at21

2

4. Rearrange equation 1. to make t the subject a

uvt

Now substitute this in equation 3 and rearrange to give :

v2 = u2 + 2as

©John Parkinson

15

USING THE EQUATIONS OF MOTION

atuv .1asuv 2.2 22 2

2

1.3 atuts

tvus 2

1.4

1. Write down the symbols of the quantities that you know

2. Write down the symbol of the quantity that you require

3. Select the equation that contains all of the symbols

in 1. and 2. abovee.g.

A stone is released from a height of 20 m above the ground. Neglecting air resistance and using the acceleration due to gravity as 9.81 ms-2, find the velocity with which the stone will hit the ground .

This must be equation 2 as it is the only one with “v”, “u”, “a” and “s” in it

v2 = u2 + 2as

v2 = 02 + 2 x 9.81 x 20

v = 392 = 19.8 m s-1

u = 0 from rest

s = 20 m

a = 9.81 ms-2

v = ?