A car starts from a point at time t=0 seconds and stops at Q. The distance x in meters covered by it in 't' seconds is given by.

$x=t^2(2-\frac{t}{3})$. The time taken by it to reach Q is :

4 seconds

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

At point $Q$, the car stops $\Rightarrow \frac{dx}{dt}=0$

$\frac{dx}{dt}=\frac{d}{dt}\left(2t^{2}-\frac{t^{3}}{3}\right)=4t-t^{2}$

$4t-t^{2}=t(4-t)=0 \;\;\Rightarrow\;\; t=0 \;\; \text{or} \;\; t=4$

Since $t=0$ is the starting point, the required time is $t=4$ seconds.

Final Answer: $4$ seconds