A child is in danger of drowning in the Merimac river. The Merimacriver has a cu

Question

A child is in danger of drowning in the Merimac river. The Merimacriver has a current of 3.7 km/hr to theeast. The child is 0.6 km from the shore and 2.6 km upstream from the dock. A rescue boat withspeed 29.1 km/hr (with respect to thewater) sets off from the dock at the optimum angle to reach thechild as fast as possible. How far from the dock does the boatreach the child?
d = 1 km    

Explanation / Answer

The child is .6 km from the shore and 2.6 km upstream Set up coordinates, you start at the origin and the child islocated at position p =2.6 i + .6 j Now, the boat is moving at a velocity RELATIVE TO THE RIVER. Thismeans that in time, any distance that the child is displacedupstream, the boat will be displaced upstream also. So, the optimumangle is right at the child or: = arctan (y/x) = arctan(.6/2.6)      :This of course, doesn'tanswer the question at all, but it's instructive to know. So, the starting distance of the child is equal to d1 = (2.6^2 +.6^2) The boat will reach the child in the time it takes to traverse thisdistance (relative to the water). d1 = 29.1* t ==> t = d1/29.1 Now that you have t you can say: t*v(river) = d2        :This isthe distance that the river carried the boat (and the child) whileit was in transit                               during time t.     vriver = 3.7 km/h Total distance is then D = d1 +d2

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