Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a constant acceleration. Both cars cross each other at time t = 0, for the first time. The maximum possible number of crossing(s) (including the crossing at t = 0) is ________.

<p>Here, we will use the concept of relative motion.</p> <p>Let initial (at $t = 0$) position of both cars is x = 0 (As they cross each other at t = 0 for Ist time)</p> <p>given : For P</p> <p>${a_P} \propto t$</p> <p>For Q,</p> <p>${a_Q} = a'$ (Let) (constant)</p> <p>$\Rightarrow {a_{{p^{(t)}}}} = kt$ where, k = constant</p> <p>$\Rightarrow {{d{v_p}(t)} \over {dt}} = kt$ (as $a = {{dv} \over {dt}}$)</p> <p>$\Rightarrow \int_0^{{v_p}(t)} {d{v_p}(t) = \int_0^t {ktdt} }$ [let p and Q both starts from rest]</p> <p>$\Rightarrow {v_p}(t) = {{k{t^2}} \over 2}$</p> <p>$\Rightarrow {{d{x_p}(t)} \over {dt}} = {{k{t^2}} \over 2}$ [As $v = {{dx} \over {dt}}$]</p> <p>$\Rightarrow \int_0^{{x_p}(t)} {d{x_{p(t)}} = \int_0^t {{{k{t^2}} \over 2}dt} }$</p> <p>$\Rightarrow {x_p}(t) = {{k{t^3}} \over 6}$ .... (1)</p> <p>Now, for Q,</p> <p>${x_Q}(t) = {1 \over 2}{a^1}{t^2}$ .... (2) [using Newton's 2nd equation of motion]</p> <p>So, relative position of P w.r.t. Q,</p> <p>${x_{P{Q^{(t)}}}} = {x_p}(t) - {x_Q}(t)$</p> <p>$\Rightarrow {x_{P{Q^{(t)}}}} = {{k{t^3}} \over 6} - {1 \over 2}{a^1}{t^2}$ (From (1) and (2))</p> <p>When both cars cross each other, $X_{PQ}=O$</p> <p>As $X_{PQ}(t)$ is a cubic polynomial, so it has maximum 3 roots.</p> <p>Hence, the maximum number of crossing = 3</p>