A car's distance is $x=t^2\left(2-\dfrac{t}{3}\right)$ metres. It starts at P ($t=0$) and stops at Q. The time taken to reach Q is

1. Velocity $v=\dfrac{dx}{dt}=4t-t^2=t(4-t)$. 2. The car stops when $v=0$: $t=0$ or $t=4$. 3. $t=0$ is at P, so it stops at Q when $t=4$ s. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.30_