# Relative speed, relative angles

1. Dec 8, 2011

### skysunsand

1. The problem statement, all variables and given/known data
A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the east at 2.0 m/s. The kayaker can paddle with a speed of 3.0 m/s.

2. Relevant equations
v= v + V

Let the earth frame be S and a frame attached to the water be S .′ Frame S′ moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames.

Solve: (a) The kayaker’s speed of 3.0 m/s is relative to the water. Since he’s being swept toward the east, he
needs to point at angle θ west of north. In frame S′, the water frame, his velocity is
v= (3.0 m/s, west of north) = (-3.0sin m/s) (3.0cos m/s)

v= (-3 sin $\theta$ + 2 ) i + (3 cos $\theta$)

In order to go straight north, v=0

so sin theta = 2/3 = 41.8 degrees

3. The attempt at a solution

The solution is above. I just don't understand what it's telling me. Why am I using cosine for Y, if cos = adjacent/hypotenuse, which is for X values?
Why is it -3 for x, instead of positive 3?
What in the world does sin theta = 2/3 have to do with this, and why does it work out like that?
If anyone can help clarify what the solution is trying to tell me, it would be very much appreciated.

2. Dec 8, 2011

### PeterO

The solution is taking North as the Y-direction, and East as the X-direction.

With the direction referrig to an angle West of North. the westerly component [x] is thus negative; and the North component [y] is positive.

EDIT: the sin theta equalling 2/3 is because the current is 2 m/s whle the kayaker is paddling at 3 m/s
also it is not -3 for X, it is -(3.sintheta)