Question

vA child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0.6 km from the shore and 2.5 km upstream from the dock. A rescue boat with speed 24.8 km/hr (with respect to the water) sets off from the dock at the optimum angle to reach the child as fast as possible. How far from the dock does the boat re?

Answer 1

Answer:

$d = 2.26 km$

Explanation:

Let the child is moving with speed same as the speed of water flow

So here the position of child with respect to flow must be zero

And if the boat start at an angle with the vertical

so its relative speed with flow of water is given as

$v_x = 24.8 sin\theta$

$v_y = 24.8 cos\theta$

now the time to reach the child is given as

$\frac{0.6}{24.8 cos\theta} = \frac{2.5}{24.8 sin\theta }$

so now we have

$\theta = 76.5 degree$

So the time to catch the child is given as

$t = \frac{0.6}{24.8 cos78.2}$

$t = 0.104 h$

So distance moved by it in 0.104 h

distance moved by the boat in upstream direction given as

$x = (24.8 sin 74.8 - 3.1)(0.104)$

$x = 2.18 km$

In y direction the displacement of boat is

$y = 0.6 km$

net displacement of the ball is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{2.18^2 + 0.6^2}$

$d = 2.26 km$