2025 DSS PRELIMS P2 Q11

2025 DSS PRELIMS P2 Q11

Secondary 4
8 marks

The diagram shows a lake (shaded) bordered by a straight level road \(AXB\). Point \(C\) is on the surface of the lake and \(B\) is the point closest to \(C\). The distance \(AB\) is 30 m and \(BC\) is 20 m.

Natalie is at point \(A\) on the road and wants to get to point \(C\) as quickly as possible. She can run at a speed of 4 m/s and swim at a speed of 2 m/s.

She realizes that in order to minimize the total time to get to \(C\), it is necessary to leave the road at some point \(X\) between \(A\) and \(B\), and then swim directly to \(C\).

Let \(AX\) be \(x\) m and \(T\) be the total time, in seconds, taken by Natalie to get from \(A\) to \(X\) and then from \(X\) to \(C\).

  1. Show that \(T=\dfrac{x}{4}+\dfrac{1}{2}\sqrt{x^2-60x+1300}\).[2]
  2. Find an expression for \(\dfrac{dT}{dx}\).[2]
  3. Find the value of \(x\) for which the total time taken is minimum. You are not required to justify that this value of \(x\) leads to the minimum total time.[4]

Solution:

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Answer:(a) \(T=\dfrac{x}{4}+\dfrac12\sqrt{x^2-60x+1300}\) (b) \(\dfrac{dT}{dx}=\dfrac14+\dfrac{2x-60}{4\sqrt{x^2-60x+1300}}\) (c) \(x=18.5\text{ m}\)

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