Topic: Engineering Mechanics
A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest path in 15 min. The velocity of the river in km/h is
- A. 3
- B. 1
- C. 4
- D. $\sqrt{41}$
Answer: A. 3
Explanation: To cross along the shortest path (straight across), the boat aims upstream at an angle $\theta$. The time taken is $15 \text{ min} = 0.25 \text{ hr}$. The effective velocity across the river is $V_{across} = \frac{\text{Distance}}{\text{Time}} = \frac{1}{0.25} = 4 \text{ km/h}$. This velocity is the component of the boat's speed perpendicular to the river flow, so $5 \cos\theta = 4$, yielding $\cos\theta = \frac{4}{5}$ and $\sin\theta = \frac{3}{5}$. The river velocity perfectly counteracts the horizontal component of the boat's velocity, so $V_{river} = 5 \sin\theta = 5 \times \frac{3}{5} = 3 \text{ km/h}$.