# Relative Velocity - boats

Hi!
I'm having trouble with some relative velocity problems.

Example A)
A boy wishes to row across a river in the shortest possible time. He can row at 2 m/s in still water and the river is flowing at 1 m/s. At which angle (theta) should he point the bow (front) of his boat.

I tried to set up a triangle and use a trig function...but I am not getting the correct answer.
The book says that the answer is 90 degrees.

I believe the relevant equation is velocity of the boat with respect to earth = velocity of the boat with respect to water + velocity of the water respect to earth

V(be)=V(bw)+V(we)
and I solve for v(bw) since they give me 2 m/s (boat with respect to earth ?) and 1 m/s (water with repsect to earth)

Then I tried to to inverse sin of (2/ sqrt(3)) but the answer is def. not 90.

alphysicist
Homework Helper
Hi crazyog,

The boat's velocity of 2 m/s is relative to the water, and the direction of the bow of the boat will be the direction of the velocity of the boat with respect to the water.

However, if I'm reading the problem correctly I don't think you actually need some of those details. The way I read it is that the boy just want to reach the other side of the river as fast as possible, and the important thing is he does not care how far downstream he happens to go along the way.

Hi!
I'm having trouble with some relative velocity problems.

Example A)
A boy wishes to row across a river in the shortest possible time. He can row at 2 m/s in still water and the river is flowing at 1 m/s. At which angle (theta) should he point the bow (front) of his boat.

I tried to set up a triangle and use a trig function...but I am not getting the correct answer.
The book says that the answer is 90 degrees.

I believe the relevant equation is velocity of the boat with respect to earth = velocity of the boat with respect to water + velocity of the water respect to earth

V(be)=V(bw)+V(we)
and I solve for v(bw) since they give me 2 m/s (boat with respect to earth ?) and 1 m/s (water with repsect to earth)

Then I tried to to inverse sin of (2/ sqrt(3)) but the answer is def. not 90.

Try considering it from the frame of the water. i.e., the water is motionless, and the bank is moving at 1 m/s.

Sheldon

Hi crazyog! Welcome to PF! ah … that's why you're getting the wrong result … "in still water" means that the 2 m/s is boat with respect to water. 