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

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

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 Click for next ►

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=16/6 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 Click for next ►

= 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 = 24 + 48 = 72m d = 72m Click for next ►

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 Click for next ►

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. Click for next ►

The Equations of Motion vf = vi + at d = ½ (vf + vi )t d = vit + ½ at2 d = vft – ½ at2 vf2 = vi2 + 2ad Click to finish