1. © 1
John Parkinson

2. Constant Velocity
Distance travelled - s
Time taken -
t
Velocity - v
s
v =
t v
s/t
Velocity = Speed in a Specified Direction
© 2
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
© 3
John Parkinson

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

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

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

7. QUESTION The graph represents the motion of a
Velocity – v/ms-1
tube train between two stations
Find
30 1. The acceleration
2. The maximum velocity
3. The retardation
20 50 80 Time – t/s 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
© 7
John Parkinson

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

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

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

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

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

13. For UNIFORM ACCELERATION
1. Distance travelled = average velocity times the time taken
Velocity
u+v v
s= t
2 u
Time
t
2. Acceleration = the change in velocity per second
v-u Rearranging
a=
t
v = u + at
© 13
John Parkinson

14. 3. Substituting equation [2] into equation [1]
u+v v = u + at
s= t
2
 u + u + at   2ut + at 
2
s=  t =  
 2   2 
1
s = ut + 2 at2
v −u
4. Rearrange equation 1. to make t the subject t =
a
Now substitute this in equation 3 and rearrange to give :
v2 = u2 + 2as
© 14
John Parkinson

15. USING THE EQUATIONS OF MOTION
1. v =u +at
1. Write down the symbols of the quantities that you know
2. v =u +2as
2 2
1
3. s = ut + at 2
2 2. Write down the symbol of the quantity that you require
1
4. s = ( u +v )t
2 3. Select the equation that contains all of the symbols
e.g. in 1. and 2. above
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
u = 0 from rest
only one with “v”, “u”, “a” and “s” in it
s = 20 m
v2 = u2 + 2as
a = 9.81 ms-2
v =? v2 = 02 + 2 x 9.81 x 20
v= 392 = 19.8 m s-1
© 15
John Parkinson