A car starts from a point P at time $t = 0$ seconds and stops at point Q. The distance $x$, in metres, 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 between P and Q.

$t = 4 \text{ s}$, $PQ = \frac{32}{3} \text{ m}$

The correct answer is Option (2) → $t = 4 \text{ s}$, $PQ = \frac{32}{3} \text{ m}$ ##

Let $v$ be the velocity of the car at $t$ seconds.

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

Therefore, $v = \frac{dx}{dt} = 4t - t^2 = t(4 - t)$

Thus, $v = 0$ gives $t = 0$ and/or $t = 4$.

Now $v = 0$ at P as well as at Q and at P, $t = 0$. So, at Q, $t = 4$. Thus, the car will reach the point Q after 4 seconds.

Also the distance travelled in 4 seconds is given by

$x\big]_{t=4} = 4^2 \left( 2 - \frac{4}{3} \right) = 16 \left( \frac{2}{3} \right) = \frac{32}{3} \text{ m}$