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(A) Let $v$ be the velocity of the car at $t$ seconds.The distance covered is given by $x = t^{2}(2 - \frac{t}{3}) = 2t^{2} - \frac{t^{3}}{3}$.The vel

Vedclass · Accepted Answer

$t=4 \, s, 	ext{ distance} = \frac{32}{3} \, m$

Vedclass · Answer

(A) Let $v$ be the velocity of the car at $t$ seconds.The distance covered is given by $x = t^{2}(2 - \frac{t}{3}) = 2t^{2} - \frac{t^{3}}{3}$.The velocity $v$ is the rate of change of distance with respect to time,so $v = \frac{dx}{dt}$.$v = \frac{d}{dt}(2t^{2} - \frac{t^{3}}{3}) = 4t - t^{2} = t(4 - t)$.The car stops at point $Q$ when the velocity $v = 0$.Setting $v = 0$,we get $t(4 - t) = 0$,which implies $t = 0$ or $t = 4$.Since the car starts at $t = 0$ at point $P$,it reaches point $Q$ at $t = 4$ seconds.The distance between $P$ and $Q$ is the value of $x$ at $t = 4$.$x(4) = 4^{2}(2 - \frac{4}{3}) = 16(\frac{6-4}{3}) = 16(\frac{2}{3}) = \frac{32}{3} \, m$.

$A$ 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 $x=t^{2}(2-\frac{t}{3})$. Find the time taken by it to reach $Q$ and also find the distance between $P$ and $Q$.

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