Question

At the initial instant, two particles are observed at different locations moving towards each other with velocities u1​ and u2​. If they are subjected to constant accelerations a1​ and a2​ in directions opposite to their initial velocities, they will meet twice. If time interval between these two meetings is Δt, find suitable expression for their initial separation

Answer 1

Let the initial separation between both the bodies be $x$

Both the bodies will cross each other twice, once during their forward motion and second time during their backward motion.

Taking the body A at rest, so the relative velocity of B with respect to A will be,

$v_{BA}=u_1+u_2$

Relative acceleration of B with respect to A,

$a_{BA}=-(a_1+a_2)$ , since acceleration is opposite to the direction of velocity

Time taken by body B to cover the distance $x$ will be,

$x=(u_1+u_2)t_1-\dfrac{1}{2}(a_1+a_2)t_1^2\longrightarrow(1)$ (this is the first time the two bodies will meet and time instant is $t=t_1$ )

Total Time taken by body B to come to rest

$v=u+at_s$

$0=(u_1+u_2)-(a_1+a_2)t_s$

$t_s=\dfrac{u_1+u_2}{a_1+a_2}$

At this time instant the body B has reached the point C from it's initial position

The difference in time interval of the two meetings will be,

$2(t_s-t_1)=\Delta t$

$t_1=t_s-\dfrac{\Delta t}{2}$

Substituting the value of $t_1$ and $t_s$ in equation (1)

$x=(u_1+u_2)\bigg(\dfrac{u_1+u_2}{a_1+a_2}-\dfrac{\Delta t}{2}\bigg)-\dfrac{1}{2}(a_1+a_2)\bigg(\dfrac{u_1+u_2}{a_1+a_2}-\dfrac{\Delta t}{2}\bigg)^2$

$x=\dfrac{(u_1+u_2)^2}{a_1+a_2}-\dfrac{\Delta t}{2}(u_1+u_2)-\dfrac{1}{2}\dfrac{(u_1+u_2)^2}{a_1+a_2}-\dfrac{(\Delta t)^2}{8}(a_1+a_2)\\\space\space\space\space\space\space\space+\dfrac{\Delta t}{2}(u_1+u_2)$

$x=\dfrac{1}{2}\dfrac{(u_1+u_2)^2}{(a_1+a_2)}-\dfrac{(\Delta t)^2(a_1+a_2)}{8}$ (Ans)