© john parkinson 1 © john parkinson 2 distance travelled - s time taken - t velocity - v v= s t v...
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©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
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DISPLACEMENT – TIME GRAPHS
Constant velocity
Displacement - s
Time - t
What will the graph look like?
GRADIENT = ?
Δt
Δs
t
sv
VELOCITY
©John Parkinson
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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
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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
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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
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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
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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
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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
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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
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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
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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 = ?