Two cars P and Q are moving on a road in the | vedclass

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(A) Let the acceleration of car $P$ be $a_P = kt$,where $k$ is a constant.Let the acceleration of car $Q$ be $a_Q = a$,where $a$ is a constant.The rel

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$3$

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(A) Let the acceleration of car $P$ be $a_P = kt$,where $k$ is a constant.Let the acceleration of car $Q$ be $a_Q = a$,where $a$ is a constant.The relative acceleration of $Q$ with respect to $P$ is $a_{QP} = a_Q - a_P = a - kt$.The relative velocity $v_{QP}$ is the integral of relative acceleration: $v_{QP} = u_{QP} + at - \frac{1}{2}kt^2$.The relative displacement $s_{QP}$ is the integral of relative velocity: $s_{QP} = u_{QP}t + \frac{1}{2}at^2 - \frac{1}{6}kt^3$.For the cars to cross,the relative displacement $s_{QP}$ must be zero.Setting $s_{QP} = 0$ gives $t(u_{QP} + \frac{1}{2}at - \frac{1}{6}kt^2) = 0$.One solution is $t = 0$ (given).The other crossings occur when the quadratic equation $u_{QP} + \frac{1}{2}at - \frac{1}{6}kt^2 = 0$ has real roots for $t > 0$.$A$ quadratic equation can have at most $2$ positive roots.Therefore,the total number of crossings is $1$ (at $t=0$) $+ 2$ (from the quadratic) $= 3$.

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