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)+√(x^2-6dx+13d^2)+9d
--------------------------------------
3v
A blue billiard ball with mass m crashes into a red billiard ball with the same mass that is at rest. The collision results in the blue billiard ball travelling with a velocity of v and the red billiard ball with a velocity of 3V. In terms of v and m determine the blue billiard ball's velocity before the collision.
2 answers
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)+√(x^2-6dx+13d^2)+9d
---------------------------------------------
3v
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)+√(x^2-6dx+13d^2)+9d
---------------------------------------------
3v
2 answers