Mathematics/Mechanics Problem

Question

I would like to ask you if anybody could help me with this problem.

So far i know that the positions where B and A have to meet are at distances L and L+2r

Answer 1

Let $t_1$ be the time taken for particle $A$ moves from $A_0$ to $H$ and let $t_2$ be the time taken for particle $A$ moves from $A_0$ to $K$, then \begin{align} d_{1A}&=v_{A0}t_1+\frac12a_At_1^2\\ l&=0+\frac12a_At_1^2\\ t_1&=\sqrt{\frac{2l}{a_A}} \end{align} and \begin{align} d_{2A}&=v_{A0}t_2+\frac12a_At_2^2\\ l+2r&=0+\frac12a_At_2^2\\ t_2&=\sqrt{\frac{2l+4r}{a_A}}. \end{align} Similarly, for particle $B$ we have \begin{align} d_{1B}&=v_{B0}t_1+\frac12at_1^2\\ \frac14\pi r&=v_{B0}t_1+\frac12at_1^2\tag1\\ \end{align} and \begin{align} d_{2B}&=v_{B0}t_2+\frac12at_2^2\\ \frac34\pi r&=v_{B0}t_2+\frac12at_2^2.\tag2\\ \end{align} Now, plug in $t_1$ to $(1)$ and $t_2$ to $(2)$, then solve the equations to obtain $v_{B0}$ and $a$.