A rower’s maximum speed in still water is 6.0 ft/s. He decides to cross a 60 ft. wide river (at his maximum speed), with an apparent heading of 30° north of east. The river’s current (relative to the ground) is flowing south at a speed of 1.0 ft/s. Find the rower’s speed and direction relative to the ground, as well as the total time it will take him to reach the other side of the river. (Please let me know if you need clarification on this – I am recalling this problem from memory, and do not have the actual problem with me at the moment)
2. Relevant equations
vR/G = VW/G + VR/W
(where R=rower, W=water, G=ground)
Law of cosines: c2 = a2 + b2 – 2abcos(θ)
Law of sines: sin(A)/a = sin(B)/b = sin(C)/c
3. The attempt at a solution
I set up the vectors in a coordinate plane. I was able to determine the angle between the rower’s apparent heading vector and the water’s vector to be 60°. Using this, I plugged it into the law of cosines formula to obtain a speed of 5.77 ft/s. I then used the law of sines (with the water’s speed and the rower’s speed of 5.77 ft/s) to obtain the angle opposite the water’s vector, which turned out to be about 8°. Subtracting this from 30°, I got about 22° north of east.
To find the time it takes to get across, I set up a right triangle, where the base represents the width of the river (60 ft), and the hypotenuse represents the distance the rower actually travels (with the angle between them 22°). Using trig, I found the hypotenuse to be about 64 ft. I then plugged this distance and the rower’s speed of 5.77 ft/s into v=d/t to solve for t, and I obtained t= ~11s.