2023 JPJC P2 Q2

2023 JPJC P2 Q2

7 marks
Prelims

In military exercises, parachutists jump from rest and their motion is tracked by sensors tagged to their bodies. One parachutist jumps vertically from a helicopter. At time \(t\) seconds, the parachutist is \(x\) metres below the helicopter with velocity \(v\)ms\(^{-1}\). It experiences a downward gravitational acceleration of \(\mathrm{9}\mathrm{.8}\)ms\(^{-1}\) and an upward acceleration \(R\) due to air resistance, where \(R<9.8\). \(R\) is directly proportional to \(v\), with a constant of proportionality \(k>0\).

The rate of change of \(v\) for the parachutist is modelled as the difference between the downward gravitational acceleration and \(R\). Write down a differential equation relating \(v\), \(t\) and \(k\), and solve this differential equation to find an expression for \(v\) in terms of \(t\) and \(k\).[6]

The terminal velocity of a falling object is the value of its velocity after a long time. State the terminal velocity of the parachutist in terms of \(k\).[1]

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Answer:\(v = \frac{9.8}{k}\left(1 - e^{-kt}\right)\); terminal velocity \(= \frac{9.8}{k}\) m/s

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