1. Deriving the equations of motion
a graphical explanation of where they come from.
By J. Jaffrey at Long Bay College, NZ

2. The 5 Variables symbol variable unit vi initial velocity ms-1 vf
final velocity a
acceleration
ms-2 d
distance m t
time s

3. Initial velocity vi = 4.0ms-1
Imagine you are cycling and reach a hill. You pedal and accelerate steadily down the hill. You start at 4.0ms-1 and after 6.0s end up at 20ms-1.
Initial velocity vi = 4.0ms-1
We can plot this motion on a graph with velocity on the vertical axis and time on the horizontal axis.
Final velocity vf = 20ms-1
Time t = 6.0s
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4. vf = vi + at Initial velocity vi = 4.0ms-1 Final velocity vf = 20ms-1
Time t = 6.0s
v (ms-1) vf 20
What is the acceleration?
Acceleration is the rate of change of velocity in ms-2. This is given by ∆v/∆t. ∆v=(vf-vi). So here: ∆v= 15
vav ∆v 10
16ms-1
Acceleration a=∆v/∆t So now we have: a=16/ a= 2.7ms-2 (2sf) 5 vi 1 2 3 4 5 6
t (s)
What is the average velocity?
The change in velocity ∆v is given by the acceleration (change in v per second) multiplied by the time (number of seconds)
So ∆v = at
Average velocity: vav = ½ (vi + vf) vav = 12ms-1
16/6 x 6 = 12ms-1
and
vf = vi + at
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5. = 48m d = vit d = vit + ½ at2 + ½ at2 = 72m v (ms-1) vf = vi + at
d = ½ (vf + vi )t 20
Can we calculate d another way? Well yes there are other ways of getting there:
If the cyclist had done a steady vi (4.0ms-1) for t (6.0s) then d would be: 15 at
vav
d = 24m 10
½ at2
The distance covered because of the increase in velocity is the average extra velocity times t or:
d = vavt 5
(½ x 16) x 6
= 48m
vit
½ x at
x t 1 2 3 4 5 6
t (s)
or ½ at2
How far did she go in the 6.0 seconds?
This is the area of the triangle under the line.
Displacement is average velocity multiplied by time so here:
So the total distance travelled is:
d = ½ (vf + vi )t
d = 12 x 6
d = vit
d = vit + ½ at2
+ ½ at2 =
= 72m
d = 72m
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6. t ½ at2 at vft vft So d = 120 – 48 = 72m d = vft – ½ at2 ½ at2 = 48m
v (ms-1)
vf = vi + at t
d = ½ (vf + vi )t 20
d = vit + ½ at2
½ at2 15 at
At vf for t seconds the cyclist covers:
vav 10
vft
20 x 6 = 120m
This is the area of the rectangle on the graph 5
vft
In fact the cyclist is slower so we must remove the area of the triangle between the line and the v axis: 1 2 3 4 5 6
t (s)
A third way to calculate d is by looking at how far the cyclist might have gone in the 6.0s of t at the final velocity of 20ms-1 vf
So d = 120 – 48 = 72m
This is rather like the last formula and is given by:
d = vft – ½ at2
½ at2
= ½ x 16 x 6
= 48m
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7. m2s-2 vf = vi + at vf = vi + at d = ½ (vf + vi )t d = vit + ½ at2
Initial velocity vi = 4.0ms-1
Final velocity vf = 20ms-1
Time t = 6.0s d = 72m
The last equation cannot be shown easily from a graph. However it can be derived using algebra.
vf = vi + at
square both sides
vf = vi + at
vf = vi + at
vf2 = (vi + at)2
expand
vf2 = vi2 + 2viat +a2t2
d = ½ (vf + vi )t
extract 2a
vf2 = vi2 + 2a ( vit + ½ at2 )
d = vit + ½ at2
But this term in brackets is simply d
so simplifying we get our final equation of motion:
d = vft – ½ at2
vf2 = vi2 + 2ad
vf2 = vi2 + 2ad
ms-2m
(ms-1)2
(ms-1)2
m2s-2
All three terms give:
We can check the units both sides are the same.
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8. The Equations of Motion
vf = vi + at
d = ½ (vf + vi )t
d = vit + ½ at2
d = vft – ½ at2
vf2 = vi2 + 2ad
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