# Easily derive kinematic eqns?

1. Feb 1, 2012

### Arijun

I'm tired of having to look up the kinematic equations every time I need to use them (I tutor lower devision students a lot). Is there an intuitive and memorable way to derive them?
As in, it's easy to derive x=x0+v0t+at2 just by integrating a constant force. Can I get the others without relying on some other equations I will inevitably forget?

2. Feb 1, 2012

### Philip Wood

The equations you refer to are, I take it, for constant acceleration.

For a start, put x0 = 0. You can always add it back in if you need it.

First approach
One basic equation is: x = (1/2)(u + v)/t [u = initial vel, v = final vel]
This embodies displacement = mean velocity $\times$ time.

The other basic one is v = u + at
This is a re-arrangement of the definition of acceleration a (if acceleration constant).

There are 5 equations in all. Each involves 4 of the 5 variables: v, u, x, t, a.

The other 3 equation are found by eliminating u, v, t in turn. If you've done it once or twice it should be quick and easy.

Second approach
Use a linear velocity-time graph running between (0, u) and (t, v). The slope (gradient) gives the acceleration, the area under the graph gives the displacement. This area is a rectangle plus a triangle. Hence the first two equations as mentioned above. Re-express the triangle area using a, and the other 3 equations will emerge.

If you use the equations often enough, you'll remember them! I do, and my memory is pretty terrible...

Note
4 of these equations relate vector quantities (or, to be fussy, components of vector quantities). The fifth equation, v2 = u2 + 2as, is different. It relates scalars (dot products). Not that you'd want to raise this with students meeting the equations for the first time!

Last edited: Feb 1, 2012
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