A swimmer wishes to cross a 500m wide river flowing at 5km/h. His speed with respect to water is 3km/h. If heads in a direction making an angle A with the flow will he reach the other shore? If no, what will be the drift, meaning what is the distance of the point he reaches from the point directly opposite him(wherehe was supposed to reach)?
Yes, he will reach the other shore (unless the angle A is 0o or 180o).
But there will be drift. The amount of drift depends on the size of angle A. If A=90o (swimming perpendicular to current) , then the amount of drift is calculated:
Time to cross = 500 m/(3000 m/hr) = 0.167 hr
Distance current carries him during that time = (5000 m/hr)*0.167 hr = 833.3 m
Suppose he chooses to make A=45o (angled upstream). Then his rate, both across the stream and in the upstream direction = (3000 m/hr)*cos45 = 2121 m/hr
[Caution: The "rate in the upstream direction" is progress through the water, not as measured by someone on the shore.]
Time to cross = 500 m/(2121 m/hr) = 0.236 hr
Distance water travels during that time = (5000 m/hr)*0.236 hr = 1179 m
But he swam upstream through that water at a rate of 2121 m/hr, so he reduced the "drift" by
(2121 m/hr)*0.236 hr = 500 m
So the actual drift is 1179 m - 500 m = 679 m.
I don't mean to suggest that an angle of 45o is the optimum angle for minimizing the drift. I suggest that you do a few more examples. Note, 45 degrees is a special case that simplified calculating the up-stream and cross-stream speed. The general case, where the angle is measured in the manner of my first paragraph:
Vcrossing = (3000 m/hr)*sinA
Vagainst_current = (3000 m/hr)*cosA
Notice that if I had answered no to the first question, that would mean he will never reach the other shore. The logic from that point would have to lead to the conclusion that the drift would be infinite.
I hope this helps,