1. MOTION
MOTION
John Parkinson

2. Velocity = Speed in a Specified Direction
Constant Velocity
Distance travelled - s
Time taken - t
Velocity - v s v = v t
s / t
Velocity = Speed in a Specified Direction
John Parkinson

3. PHYSICS BUS 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 = 25 m s-1 to the East
Velocity = ?
John Parkinson

4. DISPLACEMENT – TIME GRAPHS
Constant velocity
Displacement - s
What will the graph look like? Δs
GRADIENT = ? Δt
Time - t
VELOCITY
John Parkinson

5. What about this graph? And this graph?
Displacement - s
Time - t
What about this graph?
A body at rest
Displacement - s
Time - t Δs Δt
And this graph?
The gradient is …….?
increasing
The body must be ……..?
accelerating
John Parkinson

6. This body has a constant or uniform ………?
VELOCITY – TIME GRAPHS
Velocity - v
Time - t
This body has a constant or uniform ………? Δv Δt
acceleration
The gradient = ?
the acceleration
1 = …… ?
Uniform acceleration
Velocity - v
Time - t
2 = …… ?
Constant velocity 2 A 1
3 = …… ?
Uniform retardation [deceleration] 3
Area under the graph = A
= …….. ?
DISTANCE TRAVELLED
John Parkinson

7. QUESTION
The graph represents the motion of a tube train between two stations
Find
The acceleration
The maximum velocity
The retardation
The distance travelled
Velocity – v/ms-1
Time – t/s 30 20 50 80
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
The distance travelled = the area under the graph
=½ x 20 x x ½ x 30 x 30 = m
John Parkinson

8. s1
What will the distance – time, velocity - time and acceleration time graphs look like for this bouncing ball?
Displacement - s
Time - t s1 s2 s2
Velocity - v
Time - t
John Parkinson

9. 9.81ms-2 © John Parkinson Velocity - v Time - t Acceleration - a

10. What might this graph represent?
Velocity
What might this graph represent?
Terminal Velocity
Time
Can you draw an acceleration time graph for this motion?
Acceleration
Time
9.81 m s-1
John Parkinson

11. EQUATIONS OF UNIFORM ACCELERATION DISTANCE = VELOCITY x TIME
EQUATIONS OF MOTION
EQUATIONS OF UNIFORM ACCELERATION
For Constant Velocity
VELOCITY =
DISTANCE
TIME
If Velocity is not constant , this equation just gives the average velocity for the journey
DISTANCE = VELOCITY x TIME
John Parkinson

12. For UNIFORM ACCELERATION
EQUATIONS OF MOTION
EQUATIONS OF MOTION
For UNIFORM ACCELERATION
SYMBOLS
a = ACCELERATION
u = INITIAL VELOCITY
v = FINAL VELOCITY
s = DISTANCE TRAVELLED
t = TIME TAKEN
John Parkinson

13. For UNIFORM ACCELERATION
1. Distance travelled = average velocity times the time taken
s =
u + v 2 t
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]
u + v 2 t
v = u + at
s = ut + at2 1 2
4. Rearrange equation 1. to make t the subject
Now substitute this in equation 3 and rearrange to give :
v2 = u2 + 2as
John Parkinson

15. USING THE EQUATIONS OF MOTION
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. above
e.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 u
= 0 from rest s
= 20 m
v2 = u2 + 2as a
= 9.81 ms-2
v2 = x 9.81 x 20 v
= ?
v =
= m s-1
John Parkinson