Why Should a Boat Row at 90 Degrees to Cross a River Fastest?

AI Thread Summary
To row across a river in the shortest time, the boy should point the bow of the boat directly at 90 degrees to the riverbank, despite the river's current. The boat's speed of 2 m/s is relative to the water, while the river flows at 1 m/s. The key consideration is that the boy aims to reach the opposite bank quickly, regardless of how far downstream he drifts. Misunderstanding the velocity references led to confusion in calculations. The correct approach involves recognizing the boat's speed in still water versus the current's effect.
crazyog
Messages
50
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
crazyog said:
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
 
An answer of 90 degrees is compatible with alphysists' answer
 
Welcome to PF!

Hi crazyog! Welcome to PF! :smile:
crazyog said:
… they give me 2 m/s (boat with respect to Earth ?)

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. :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top