# Relative Velocity - Canoe vs. River

1. Aug 16, 2014

### killaI9BI

1. The problem statement, all variables and given/known data

A canoeist wants to travel straight across a river that is 0.10km wide. However, there is a strong current moving downstream with a velocity of 3.0km/h. The canoeist can maintain a velocity relative to the water of 4.0km/h.

a) In what direction should the canoeist head to arrive at a position on the other shore directly opposite to his starting position?

b) How long with the trip take him?

2. Relevant equations

a2 + b2 = c2

sin a = (sin b/b) X a

3. The attempt at a solution

a) r2 = 42 + 32 = 25
r = 5km
sin θ = (sin90/5)X 3
sin θ = 0.6
θ = 37

He should aim 37° upstream, from the destination reference.

The book's answer is: He should aim upstream at an angle 41° with respect to the riverbank.

b) cos 37 = 0.1/hyp
hyp = 0.1/0.79
hyp = 0.125
0.125/4 = 0.03 = 1.9min

The book's answer is 2.3 min.

Any help that you can provide is greatly appreciated!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 16, 2014

### Staff: Mentor

First things first. Draw yourself a vector diagram. Yes, a right triangle is involved, but what is the hypotenuse?

3. Aug 16, 2014

### killaI9BI

Thank you.

I've done that and thought the hypotenuse was calculated from the angle as well as the length of the adjacent side:
cos 37 = 0.1/hyp
hyp = 0.1/0.79
hyp = 0.125
0.125/4 = 0.03 = 1.9min

is this the wrong way? I figured my answer must be wrong because the angle that I calculated for question a) is different than the book's answer.

4. Aug 16, 2014

### Nathanael

You want to find the angle in which he will have no upstream/downstream velocity.

This means the up-stream-component of his velocity* must equal the downstream velocity of the water.

At what angle does this happen?

*(his velocity relative to the water)

5. Aug 16, 2014

### Staff: Mentor

You are mixing up which side is the hypotenuse. (You won't calculate the hypotenuse--it's given.)

Draw the triangle representing this vector addition:
V(water/shore) + V(boat/water) = V(boat/shore)

This forms a right triangle. But the trick is correctly identifying the sides.

6. Aug 17, 2014

### killaI9BI

I'm so confused.
vws=3km/h
vcw=4km/h

32+42=52
vcs=5km/h

My brain is mush. That number doesn't seem right.

I don't understand how this question is messing me up so much.

7. Aug 17, 2014

### phinds

And we likely won't understand how to help you unless you show your work. Your work INCLUDES the vector diagram, so draw it and show it.

8. Aug 17, 2014

### Staff: Mentor

It's not right. Show us that vector diagram.

Your analysis assumes that vws and vcw are perpendicular. They are not.

9. Aug 17, 2014

### killaI9BI

I apologize

10. Aug 17, 2014

### killaI9BI

ohhhh!

I will redo my work considering that but I don't think I know where to start.

Last edited: Aug 17, 2014
11. Aug 17, 2014

### Staff: Mentor

Finally, a diagram!

You are told that the canoeist wants to go directly across to the other shore, so Vcs must be straight across. The angle that Vcw makes is what you need to find. (Using a corrected diagram.)

12. Aug 17, 2014

### killaI9BI

so vcw=4
and vws=3
so vcs=2.65

finally!

so (0.1/2.65)x60 = 2.3 min

and sin Θ = (sin 90/4) X 2.65
= 41°

Hooray!!!!

thank you very much!!!

13. Aug 17, 2014

### Staff: Mentor

Yay!

14. Aug 17, 2014

### phinds

Nah, you did fine. The trick to these, as I think you now see, is to get the vector diagram right. The math is trivial but you need to be sure you are actually solving the right problem.