Answer:
Let they move with uniform speeds \[{{V}_{1}}\]and\[{{V}_{2}}.\]
For case CA: \[\frac{s+L}{{{V}_{1}}}-\frac{x+L}{{{V}_{2}}}=t\]
\[\Rightarrow \frac{1}{{{V}_{1}}}-\frac{1}{{{V}_{2}}}=\frac{t}{x+L}\] ……(1)
For case CB: \[\frac{L}{{{V}_{1}}}-\frac{L}{{{V}_{2}}}=T\]
\[\Rightarrow \frac{1}{{{V}_{1}}}-\frac{1}{{{V}_{2}}}=\frac{T}{L}\] ………(2)
Equating (1) and (2) we get, \[x=L\left( \frac{t-T}{T} \right)\]
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