Question

∆x=vt-1/2at^2 rearrange this equation to find acceleration?

Answer 1

Hey MATE!

Δx is change in position or we represent it as dv.

Therefor,

$dv \: = \: vt \: - \frac{1}{2} a {t}^{2} \\ \\ = > \frac{dv}{dt} = v \: - \frac{1}{2} a \: 2t \\ \\ = > v \: - \: at \:$

which gives us acceleration.

Hope it helps

Hakuna Matata :))