Question

A car starts from a point P at time t = 0 seconds and stops at a point Q. The distance x, in meters, covered by it, in t seconds is given by $x=t^2\left(2-\frac{t}{3}\right).$ Find the time taken by it to reach Q and also find distance PQ.

Answer 1

$$\begin{gathered} \mathrm{Distance}PQ\mathrm{is}\mathrm{given}\mathrm{as}x=t^2\left(2-\frac{t}{3}\right) \\ \Rightarrow x=2t^2-\frac{t^3}{3} \\ \Rightarrow \frac{dx}{dt}=4t-t^2 \\ \Rightarrow \frac{dx}{dt}=t\left(4-t\right) \\ \mathrm{Now},\mathrm{the}\mathrm{car}\mathrm{stops}\mathrm{at}\mathrm{point}Q\mathrm{so},\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{car}\mathrm{at}\mathrm{point}Q\mathrm{is}0. \\ \Rightarrow t\left(4-t\right)=0 \\ \Rightarrow t=0,t=4 \\ \mathrm{So},\mathrm{the}\mathrm{car}\mathrm{takes}4\mathrm{s}\mathrm{to}\mathrm{reach}\mathrm{point}Q. \\ \mathrm{Now},\text{Distance covered by car i.e.}\mathit{\text{PQ}}\text{=}\mathit{\text{x}}\text{=}{t}^2\left(2-\frac{t}{3}\right) \\ \Rightarrow x=4^2\left(2-\frac{4}{3}\right) \\ \Rightarrow x=16\times \frac{2}{3} \\ \Rightarrow x=\frac{32}{3}\mathrm{m} \end{gathered}$$