2022 HCI P1 Q12

2022 HCI P1 Q12

14 marks
Prelims

A cargo drone is used to unload First Aid kit at an accident location in a remote mountain area. The First Aid kit is unloaded vertically through air. The speed of the kit, \(v\)ms\(^{-1}\), is the rate of change of distance,\(x\) m, of the kit measured vertically away from the drone with respect to time, \(t\) seconds.

  1. Write down a differential equation relating \(v\), \(x\) and \(t\).[1]

The motion of the First Aid kit is modelled by the differential equation

\(\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{t}^{2}}}+\alpha {{\left( \frac{\mathrm{d}x}{\mathrm{d}t} \right)}^{2}}=10\), \(\left( \mathrm{A} \right)\)

where \(\alpha \) is a constant.

  1. By using the result in part (i), show that the differential equation \(\left( \mathrm{A} \right)\) can be expressed as
    \(\frac{\mathrm{d}v}{\mathrm{d}t}=10-\alpha {{v}^{2}}\)[1]

It is given that when \(t=0\), \(v=0\). It is also given that \(\frac{\mathrm{d}v}{\mathrm{d}t}=4.375\) when \(v=1.5\).

  1. Find the value of \(\alpha \). By solving the differential equation in part (ii), show that
    \(v=\frac{k-k{{\mathrm{e}}^{-10t}}}{m+{{\mathrm{e}}^{-10t}}}\),
    where \(k\) and \(m\) are constants to be determined.[4]
  2. Sketch the graph of \(v\) against \(t\) and describe the behaviour of \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{t}^{2}}}\) in the long run.[4]
  3. Show that the area under the graph in part (iv) bounded by the lines \(t=0\) and \(t=T\) can be expressed as \(\frac{2}{5}\ln \left( \frac{{{\mathrm{e}}^{\beta \,T}}+{{\mathrm{e}}^{-\beta \,T}}}{2} \right)\), where \(\beta \) is a positive integer to be determined.[3]
  4. What can be said about \(\frac{2}{5}\ln \left( \frac{{{\mathrm{e}}^{\beta \,T}}+{{\mathrm{e}}^{-\beta \,T}}}{2} \right)\) in the context of this question?[1]

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Answer:(i) \(v=\frac{\mathrm{d}x}{\mathrm{d}t}\). (ii) \(\frac{\mathrm{d}v}{\mathrm{d}t}=10-\alpha v^2\). (iii) \(\alpha=2.5\), \(v=\frac{2-2e^{-10t}}{1+e^{-10t}}\), \(k=2\), and \(m=1\). (iv) \(v\to2\text{ m s}^{-1}\) and \(\frac{\mathrm{d}v}{\mathrm{d}t}\to0\text{ m s}^{-2}\). (v) \(\beta=5\). (vi) The expression is the First Aid kit's vertical distance travelled from time \(0\) to time \(T\).

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