Assuming the direction of the water current is along the positive \(x\)-axis and the direction of the swimmer is along the positive \(y\)-axis.
The magnitude of the water current \(=1.25\) km h\(^{-1}\)
The magnitude of the swimmer \(=2.5\) km h\(^{-1}\)
Let \(v\) km h\(^{-1}\) represent the resultant velocity of the swimmer.
![A triangle illustration displaying its area and height, exemplifying a vector representation in triangular form.](https://question.pandai.org/uploads/image/246491/KSSM-F4-AM-08-01__Vectors_11_Ammend.png)
Hence, \(v=1.25 \utilde{i} + 2.5 \utilde{j}\).
The magnitude of the resultant velocity of the swimmer is:
\(\begin{aligned} |\utilde{v}|&= \sqrt{1.25^2+2.5^2} \\ &=2.795 \text{ km h}^{-1}. \end{aligned}\)
Since the width of the river is \(200 \text{ m}= 0.2 \text{ km}\),
hence, \(\begin{aligned} \text{Time taken}&=\dfrac{0.2}{2.795} \\\\ &=0.07156 \,\text{ hour}. \end{aligned}\)
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