Question

A particle moves along the x-axis obeying the equation $x=t(t-1)(t-2)$, where $x$ is in meter and $t$ is in second
Find the acceleration of the particle when its velocity is zero.

Answer 1

$x=t[t-1][t-2]=(t^2-t)(t-2)$

$=t^3-2t^2-t^2+2t=t^3-3t^2+2t$

$v=\frac{dx}{dt}=3t^2-6t+2$

$a=6t-6=6\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)-6=\frac{6}{\sqrt{3}}=-2\sqrt{3}m{s}^{-2}$ and

$a=6\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)-6=\frac{6}{\sqrt{3}}=2\sqrt{3}m{s}^{-2}$