## Motion with Constant Acceleration

What sort of motion does this equation describe? x(t) = x0+v0t + 1/2 a t2 (1)

v(t) = v0+a0t (2)

a(t) = a0 (3)

This is motion with constant acceleration (a(t) = constant).

Equations (1) and (2) are important equations to remember, as problems with constant acceleration are very common in this course – for example, we will shortly see them when we start to talk about projectile motion in two dimension. There is a third important equation, which can be found thusly:

Rearranging (2) --> t=(v-v0)/a0 (4)

Substituting (4) into (1) yields:

x=x0 + v0(v-v0)/a0 + 1/2 a0(v-v0)2/a02

x=x0+vv0/a0 – v02/a0 +1/2 v2/a0 –vv0/a0 + v02/a0

x=x0+v2/2a0 – v02/2a0

or:

v2-v02=2a0(x-x0) (5)
There is a faster way to derive (5) if we are willing to make a few assumptions/guesses.

Clearly, x=x0+vaveraget If we assume that vaverage = (v+v0)/2 (which is not generally true, but it is easy to show that it is true for constant acceleration), then we have:

x-x0=vaverage(v-v0)/a0

= (v+v0)(v-v0)/2a0 = (v2-v02)/2a0

as before (here we have used (4) again for ‘t’). Either way, we have a technique that relates the initial and final velocities and positions without needing the initial or final times. There are many problems that this is useful for.