Solution:
The second body travels \(s/2\) at constant speed \(2v\) in time \(t = s/(4v)\). For the first body: \(s/2 = vt + \frac{1}{2}at^2\). Substituting \(t\) gives \(s/2 = s/4 + a s^2 / (32v^2)\), which simplifies to \(a = 8v^2/s\).
Two bodies separated by a distance of \('s'\) start moving towards each other with speeds of \(v\) and \(2v\) respectively. The uniform acceleration with which the first body should move so that they meet at the middle is:
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