## Motion Question

1.Jack is strolling along the bank of a 1-km wide River when the most beautiful girl materialises on the shore directly opposite him. Fearing she will disappear before he has a chance to establish face-to-face communication, he quickly devises a plan to reach the opposite shore in the shortest possible time. There is a rowboat breached on the shore right infront of him. He knows he can row at a speed of 6 km/h in still water and the current has a speed of 3 km/h. He must reach the opposite side of the river in the shortest time. His path includes a diagonal trip across the river followed by a sprint along the opposite shore to reach Jill. Jack has a standard sprinting speed of 10 km/h. What direction did he head his boat.

P.S Im still looking for number 2 on my other post. I thought I had the answer but it was wrong.
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 Recognitions: Gold Member Science Advisor Staff Emeritus Technical point: you say in your post "He must reach the opposite side of the river in the shortest time." Taken literally, that would mean that it doesn't matter WHERE on the opposite shore he winds up and his running speed after he gets there is irrelevant. I am going to take it that Jill will remain where she is and Jack wants to get to her position in the shortest time. Let theta be the angle at which he rows relative to the straight line across the river (directly across would be theta= 0, aiming downstream positive, upstream negative). Since his speed in still water is 6 km/hr, his "velocity vector" would be (6 sin(theta), 6 cos(theta)). Since the river is flowing at 3 km/, its velocity vector is (3, 0) and Jack's "velocity made good" is (6 sin(theta)+ 3, 6 cos(theta)). He will cross the river in time t1 if 6 cos(theta)t1= 1 or t1= 1/(6 cos(theta)). At that time, his position on the far shore will be (6 sin(theta)+ 3)sin(theta), positive if downstream from Jill, negative if upstream. In any case, he now has that distance to run at 10 km/h That will require time t2= 10(6 sin(theta)+ 3)sin(theta). You want to minimize t1+ t2= 1/(6 cos(theta))+10(6 sin(theta)+ 3)sin(theta). Differentiate with respect to theta and set equal to 0.