Question

A canoe has a velocity of 0.550 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.490 m/s east relative to the earth.

a. Find the magnitude of the velocity of the canoe relative to the river.

b. Find the direction of the velocity of the canoe relative to the river.

 

Answer 1

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

The answer is 0.4018 m/s.

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 °

The answer is 75.43 °.