A car starts from a point P at time t = 0 seconds and stop at Q. The distance x, in metres, covered by it in t seconds is given by $x(t) =t^3\left(3-\frac{t^2}{5}\right)$. The distance between P and Q is :

$\frac{162}{5}m$

The correct answer is Option (2) → $\frac{162}{5}m$

$x=t^3\left(3-\frac{t^2}{5}\right)$

$\frac{dx}{dt}$= velocity = v = $3t^2\left(3-\frac{t^2}{5}\right)-\frac{2t}{5}×t^3$

at Q (v = 0) as car stops

so $3t^2\left(3-\frac{t^2}{5}\right)-\frac{2t^4}{5}=0$

$3\left(3-\frac{t^2}{5}\right)-\frac{2t^2}{5}=0$

so $9-\frac{3t^2}{5}-\frac{2t^2}{5}=0⇒t^2=9$

$t=3s$

at $x(3)=3^3(3-\frac{9}{5})=27(\frac{6}{5})=\frac{162}{5}m$