Relative Velocity - Canoe vs. River

[killaI9BI]

Homework Statement

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?

Homework Equations

a² + b² = c²

sin a = (sin b/b) X a

cos θ = adjacent/hypotenuse

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!

 

[Doc Al]

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

 

[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.

 

[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)

 

[Doc Al]

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.

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.

 

[killaI9BI]

I'm so confused.
v_ws=3km/h
v_cw=4km/h

3²+4²=5²
v_cs=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.

 

[phinds]

I'm so confused.
v_ws=3km/h
v_cw=4km/h

3²+4²=5²
v_cs=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.

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.

 

[Doc Al]

I'm so confused.
v_ws=3km/h
v_cw=4km/h

3²+4²=5²
v_cs=5km/h

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

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

Your analysis assumes that v_ws and v_cw are perpendicular. They are not.

 

[killaI9BI]

I apologize

 

[killaI9BI]

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

Your analysis assumes that v_ws and v_cw are perpendicular. They are not.

ohhhh!

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

 

[Doc Al]

I apologize

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.)

 

[killaI9BI]

talking to me about this is harder than teaching a rock how to swim!

so v_cw=4
and v_ws=3
so v_cs=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!

 

[Doc Al]

Yay!

 

[phinds]

talking to me about this is harder than teaching a rock how to swim!

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.