AB and CD are parallel paths of equal length $$d$$. Two persons P and Q start simultaneously from A and C and walk towards B and D with speeds $$V_1$$ and $$V_2$$. On reaching the destination they reverse their directions. If $$V_1$$ is greater than $$V_2$$ where do they cross?
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PLACEHOLDER ANSWER - the answer is an expression. P reaches B first and turns back, so at time $$t$$ after that P is at $$2d - V_1 t$$ from the start while Q is at $$V_2 t$$. They cross when $$V_2 t = 2d - V_1 t$$, i.e. $$t = \frac{2d}{V_1 + V_2}$$. This is at a distance $$\frac{2dV_2}{V_1 + V_2}$$ from A and C (or $$\frac{d(V_1 - V_2)}{V_1 + V_2}$$ from B and D).