Question 105: Motion in 1D

Question

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 ________.

Accepted Answer

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

Motion in 1D

Integer / Numeric Answer Type