2026 ASRJC P1 Q12

2026 ASRJC P1 Q12

Junior College 2
13 marks

An ocean-monitoring sensor stores electrical energy generated from wave motion. Let \(E\) joules be the amount of energy stored \(t\) hours after the sensor is activated.

At the time of activation,

  • wave motion supplies energy at a rate of \(\displaystyle \frac{12}{t+1}\) joules per hour;
  • energy is consumed by the sensor at a rate of \(\displaystyle \frac{3E^2}{16(t+1)}\) joules per hour.

At the time of activation, the sensor already contains 4 joules of stored energy.

  1. Show that the rate of change of energy per hour is given by \(\displaystyle \frac{\mathrm{d}E}{\mathrm{d}t}=\frac{3(64-E^2)}{16(t+1)}\).[1]
  2. Hence, show that the energy stored is given in the form \(\displaystyle E=8-\frac{A}{3(t+1)^3+1}\), where \(A\) is a positive constant.[6]
  3. Deduce the amount of energy stored in the sensor in the long run.[1]
  4. Sketch the graph of \(E\) against \(t\) and state the range of \(E\).[3]
  5. On average, a town requires \(9\times10^8\) joules of energy per hour. If the plan is to generate 2% of this energy using \(3\times10^6\) of these sensors, how long after the sensors are activated will it take for the sensors to store enough energy to meet this target? Give your answer to the nearest minute.[2]

Solution:

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Answer:(i) \(\frac{\mathrm{d}E}{\mathrm{d}t}=\frac{3(64-E^2)}{16(t+1)}\) (ii) \(E=8-\frac{16}{3(t+1)^3+1}\), \(A=16\) (iii) \(8\text{ J}\) (iv) increasing from \((0,4)\) to asymptote \(E=8\), range \([4,8)\) (v) \(20\) minutes

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