N2024 P1 Q11

A helicopter is hovering at a great height above the ground. A parachutist drops from the helicopter and his parachute opens. The parachutist falls vertically. At time \(t\) seconds after leaving the helicopter he has fallen a distance of \(x\) metres. At this time he has velocity \(v\mathrm{~m s}^{-1}\). The motion of the parachutist is modelled by the differential equation

\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \alpha \frac{\mathrm{d}x}{\mathrm{d}t} = 9.8,\)

where \(\alpha\) is a constant.

1. 1. Use the equation \(v = \frac{\mathrm{d}x}{\mathrm{d}t}\) to write down a differential equation in \(v\) and \(t\).[1]
   2. Given that \(\frac{\mathrm{d}v}{\mathrm{d}t} = 5\) when \(v = 2.4\), find the value of \(\alpha\).[1]
2. It is given that the initial velocity of the parachutist is zero. Show that \(v = A(1 - \mathrm{e}^{Bt})\), where \(A\) and \(B\) are constants to be found.[5]
3. The helicopter is at a height of \(2000\mathrm{~m}\) above the ground. Find the time taken for the parachutist to reach the ground.[5]