Need some help with two Dimensional velocity Vectors

[Giu1iano]

Mod note: Moved from a technical section, so missing the homework template.
Hi, I'm struggling to understand this problem. I would normal just wait for class but I'm away traveling for work.

The question is as follows.
A man has a boat that can travel at 15.0 m/s relative to the water. He starts out from the south shore of a river and plans to cross so that he lands directly across the river from his starting point. The river has a velocity of 8.00 m/s East relative to the shore.

a) at what angle should he point his boat?
b) what will be his velocity relative to the shore?
c) if the river is 300 meters wide, how long will it take him to cross?

Answers: a) [N32°W] b) 12.7 m/s [N] c) 23.6s

For a) I used Tan=8/15, Tan=28° or [N28°W] (which is wrong)
b) I used a^2+b^2=c^2 to solve the resultant Lane. 8^2+15^2=289, √289=17 or 17 m/s[N] (which is wrong)
c) I used time=displacement/velocity, t=300m/17m/s, t=17.6s.

I missed the lecture for this class because of work, and my teachers notes confuse me. I'd appreciate any help!

Thank you :)

 

[berkeman]

For a) I used Tan=8/15

Welcome to the PF.

The above is not right... If the boat's speed is 15, that will be the hypotenuse. The tan function is for the two opposite sides. What would the correct trig function be?

 

[Giu1iano]

I'm sorry to say but you're not right either. The speed of the boat is 15 m/s. If the boat starts from the south shore of the river and plans to cross so that he lands directly across from where he started that would put him traveling North. 15 is adjacent and 8 is the opposite.

 

[berkeman]

I'm sorry to say but you're not right either. The speed of the boat is 15 m/s. If the boat starts from the south shore of the river and plans to cross so that he lands directly across from where he started that would put him traveling North. 15 is adjacent and 8 is the opposite.

No, I don't think so. The diagram I drew has a triangle with the short 8m/s side on the bottom (north points up), the hypoteneuse is 15m/s, and the left vertical side is the river crossing speed which is less than 15 but more than 8. The angle at the top of the triangle is the same as the angle of the 15m/s vector to the left of the vertical. The angle at the top of the triangle has 15m/s as the hypotenuse and 8m/s as the base at the bottom (opposite side)... I do get the same answers as you have posted, BTW.

 

[Giu1iano]

No, I don't think so. The diagram I drew has a triangle with the short 8m/s side on the bottom (north points up), the hypoteneuse is 15m/s, and the left vertical side is the river crossing speed which is less than 15 but more than 8. The angle at the top of the triangle is the same as the angle of the 15m/s vector to the left of the vertical. The angle at the top of the triangle has 15m/s as the hypotenuse and 8m/s as the base at the bottom (opposite side)... I do get the same answers as you have posted, BTW.

Mmk, let me try

 

[SammyS]

The hypotenuse does correspond to the velocity of 15 m/s. His velocity (not speed) relative to the water must have a component of 8m/s toward the West to counter the velocity of the river.

If the river had zero velocity, it would take 20 seconds to cross. It takes longer if the river is flowing & he wants to go directly North.

 

[Giu1iano]

Thank for helping me realize my mistake!