What about x = x0+v0t + 1/2 a t?

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+v_averaget 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=v_average(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.