Question

Two particles P and Q move in a straight line AB towards each other. P starts from A with velocity `u_(1)` and an acceleration `a_(1)` Q starts they pass each other at the midpoint of AB and arrive at the other ends of AB with equal velocities.
A. they meet at midpoint at time `t=(2(u_(2)-u_(1)))/((a_(1)-a_(2))`
B. The length of path specified i.e., `AB` is `l=(4(u_(2)-u_(1))(a_(1)u_(2)-a_(2)u_(1)))/((a_(1)-a_(2))^(2))`
C. They reach the other ends of AB with equal velocities if `(u_(2)+u_(1))(a_(1)-a_(2))=8(a_(1)u_(2)-a_(2)u_(1))`
D. They reach the other ends of AB with equal velocities if `(u_(2)-u_(1))(a_(1)+a_(2))=8(a_(2)u_(1)-a_(1)u_(2))`

Answer 1

Correct Answer - A::B::C

`(1)/(2)=u_(1)t+(1)/(2)a_(1)t^(2)` …(1) and `-(1)/(2)=-u_(1)t+(1)/(2)(-a_(2))t^(2)`
`implies(1)/(2)=u_(2)t+(1)/(2)a_(2)t^(2)` ..(2)
Substracting (1) and (2) or we get `t=2((u_(2)-u_(1))/(a_(1)+a_(2)))` ..(3)
Substituting (3) in (1) or (2) and rearranging we get
`1=(4(u_(2)-u_(1)))/(a_(1)+a_(2))^(2)(a_(1)u_(2)-a_(2)u_(1))` ..(4)
since the particle P & Q reach the other ends of A and B with equal velocities say v
For particle P `v^(2)-u_(1)^(2)=2a_(2)1` ..(5)
for particle Q `v^(2)-u_(2)^(2)=2a_(2)1` ..(6)
Substracting and then substituting value of 1 and rearranging we get
`(u_(2)+u_(1))(a_(1)-a_(2))=8(a_(1)u_(2)-a_(2)u_(1))`