Let the man run to point P, which is x meters upstream from the line AB. He swims across, being swept y meters downstream, and then runs to B.
We know that he is in the water for d/(v/3) = 3d/v seconds, so
y = 3d/v * v = 3d.
His path is thus
√(d^2+x^2) meters on land,
d meters in the water,
√((2d)^2+(x-3d)^2) meters on land to B
Now, dividing distance by speed, his time to travel is
t = 1/(3v) (√(d^2+x^2)+√((2d)^2+(x-3d)^2)) + 3d/v
=
√(d^2+x^2)+√(4d^2+(x-3d)^2)+9d
-----------------------------------------------
3v
Now, to find minimum time, we need dt/dx = 0.
Once you get dt/dx and simplify things a bit, you need to solve
x√(d^2+x^2)+(x-3d)√(4d^2+(x-3d)^2) = 0
I get x = 5d/3
A river flows at the speed of v from west to east. How should a man who is at a point A of the southern bank of the river, which is d meters away from the river, and wants to reach a point B on the northern bank of the river, which is 2d meters away from the river, choose his path in order to reach the destination at the shortest time? Assume the width of the river to be d, the man to run at a speed of 3v and swim at the speed of v/3 in still water, and the points A and B on the same meridian.
1 answer
1 answer