**A Canoe on a River**

Part A = 0.4018 m/s

Part B = 75.43 °

**Solution Below:**

Part A

This is kind of a confusing problem. In case Mastering Physics doesn’t give you the angle for south east, it’s 315 °. Start by finding the components (x and y) of the canoe’s velocity with respect to the river. This is:

v(canoe, river) = v(canoe, earth) – v(river, earth)

So find the velocity for the x and y components. We’ll do x first:

Note that the angle for the canoe vs the earth is 315 ° (south east), but since the river is flowing directly east, the angle for the river vs the earth is 180 °.

vx(canoe, river) = vx(canoe, earth) – vx(river, earth)

vx(canoe, river) = 0.55 * cos(315) + 0.49 * cos(180)

vx(canoe, river) = -0.1011

Now find the y component of the velocity:

vy(canoe, river) = vx(canoe, earth) – vx(river, earth)

vy(canoe, river) = 0.55 * sin(315) + 0.49 * sin(180)

vy(canoe, river) = -0.3889

Now find the velocity by using the Pythagorean theorem:

v = sqrt(x^2 + y^2)

v = sqrt(-0.1011^2 + -0.3889^2)

v = 0.4018 m/s

0.4018 m/s

Part B

Express your answer as an angle measured south of west.

Just use the component velocities (x and y) that you found in Part A. Use the atan to get the angle:

θ = atan(y / x)

θ = atan(-0.3889 / -0.1011)

θ = 75.43 °

75.43 °

0.55 * cos(315) – 0.49 * cos(180)=.8789 in degree mode on my calculator. you have -0.1011 How are you coming to this?

You’re right, it should be 0.55 * cos(315) + 0.49 * cos(180), instead. This will give you the -0.1011