Out and back under a falling acceleration

Out and back under a falling acceleration

IP 4
9 marks

A particle starts from rest at a fixed point \(O\) and moves in a straight line. Its acceleration \(t\) seconds later is \(a=k-2t\) m/s\(^2\), where \(k\) is a positive constant.

  1. Find expressions for the velocity and the displacement from \(O\) in terms of \(k\) and \(t\).[2]
  2. Show that the particle returns to \(O\) when \(t=\dfrac{3k}{2}\).[2]
  3. Show that the greatest distance of the particle from \(O\) before it returns is \(\dfrac{k^3}{6}\) m, justifying that the particle changes direction only once before returning.[3]
  4. The particle travels a total distance of \(72\) m from leaving \(O\) until it first returns. Find \(k\).[2]

Solution:

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Answer:(a) \(v=kt-t^2\), \(s=\dfrac{kt^2}{2}-\dfrac{t^3}{3}\) (b) shown (c) the only reversal is at \(t=k\), where \(s=\dfrac{k^3}{6}\) (d) \(k=6\)

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