Acceleration for Midpoint Meeting – Rankers Physics

Relative Motion in One Dimension: Practice Problem & Solution

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:
\(\frac{v^2}{s}\)
\(\frac{v^2}{2s}\)
\(\frac{8v^2}{s}\)
\(\frac{4v^2}{s}\)

Solution Explained:

To solve this problem, we apply the core principles of Relative Motion in One Dimension. Understanding the underlying formula is key to arriving at the correct answer below:

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 \(a = 8v^2/s\).

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