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JEE PYQ: Motion in a Straight Line - Question ID 5556ba9f432e (JEE Main 2025)

ID: 5556ba9f432eJEE Main 2025Numerical Value

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 ________.

Your Answer

Step-by-step Explanation

Here, we will use the concept of relative motion.

Let initial (at $t = 0$ ) position of both cars is x = 0 (As they cross each other at t = 0 for Ist time)

given : For P

${a_P} \propto t$

For Q,

${a_Q} = a'$ (Let) (constant)

$\Rightarrow {a_{{p^{(t)}}}} = kt$ where, k = constant

$\Rightarrow {{d{v_p}(t)} \over {dt}} = kt$ (as $a = {{dv} \over {dt}}$ )

$\Rightarrow \int_0^{{v_p}(t)} {d{v_p}(t) = \int_0^t {ktdt} }$ [let p and Q both starts from rest]

$\Rightarrow {v_p}(t) = {{k{t^2}} \over 2}$

$\Rightarrow {{d{x_p}(t)} \over {dt}} = {{k{t^2}} \over 2}$ [As $v = {{dx} \over {dt}}$ ]

$\Rightarrow \int_0^{{x_p}(t)} {d{x_{p(t)}} = \int_0^t {{{k{t^2}} \over 2}dt} }$

$\Rightarrow {x_p}(t) = {{k{t^3}} \over 6}$ .... (1)

Now, for Q,

${x_Q}(t) = {1 \over 2}{a^1}{t^2}$ .... (2) [using Newton's 2nd equation of motion]

So, relative position of P w.r.t. Q,

${x_{P{Q^{(t)}}}} = {x_p}(t) - {x_Q}(t)$

$\Rightarrow {x_{P{Q^{(t)}}}} = {{k{t^3}} \over 6} - {1 \over 2}{a^1}{t^2}$ (From (1) and (2))

When both cars cross each other, $X_{PQ}=O$

As $X_{PQ}(t)$ is a cubic polynomial, so it has maximum 3 roots.

Hence, the maximum number of crossing = 3

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