Question

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

Answer 1

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 $\tan \theta =\frac{3}{4}$

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

Time taken is $20\times \frac{3}{5}t=0.6$

$t=\frac{0.6}{12}=0.05h=(0.05\times 60)=3min$