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Question

A child in danger of drowning in a river is being carried downstream by a current that flows uniformly at a speed of $$2.5\ km/h$$. The child is $$0.6\ km$$ from shore and $$0.8\ km$$ upstream of a boat landing when a rescue boat sets out. If the boat proceeds at its maximum speed of $$20\ km/h$$ with respect to the water, how long (in $$min$$)does it take the boat to reach the child? 


Solution

The boat moves on a path that makes an angle of $$\theta$$ with respect to the shore line.
Call the direction of river flow the x axis and the direction across the river as the y axis.
Then the component of motion of the boat along the x axis is $$20 \cos\theta$$ and y component $$20 \sin \theta$$

Therefore $$20 \cos\theta \times t = 0.8$$
and $$20 \sin \theta \times t = 0.6$$
Divide both the equation we get $$\tan \theta = \dfrac{3}{4}$$

Therefore, $$\sin \theta = \dfrac{3}{5}$$

Time taken is $$20 \times \dfrac{3}{5}t = 0.6$$
$$t = \dfrac{0.6}{12} = 0.05\ h = (0.05 \times 60) = 3\  min$$

Physics

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