eye of hurricane magnitude?

Question

The eye of a hurricane passes over Grand Bahama Island. It is moving in a direction 60.0° north of west with a speed of v1 = 37.0 km/h. Exactly three hours later, the course of the hurricane shifts due north, and its speed slows to v2 = 27.0 km/h, as shown in Figure 3-30. How far from Grand Bahama, in kilometers, is the hurricane 4.50 h after it passes over the island?

** for my answer i got 118 km the first time, the second time i got 104 km using tangent and finding the angle which i got 20 degrees and then used that to find the hypoteneuse.

Answer 1

Here's what I get. I'm using angles counterclockwise relative to +x or East. To add two paths defined by velocities in polar coordinates (magnitude and angle) and times, you have to convert each velocity to a vector, then scale the vectors by their times to obtain displacements, then sum the displacements and convert the result back to polar.
Convert 1st part velocity to vector, then get displacement.
mag=37 kph, angle=120
v1 = (-18.5, 32.043) km/h
x1 = v1*3 h = (-55.5, 96.13) km
Convert 2nd part velocity to vector, then get displacement.
mag=27 kph, angle=90
v2 = (0, 27) km/h
x2 = v2*1.5 h = (0, 40.5) km
Sum the two displacements
xt = x1+x2 = (-55.5, 136.63) km
Convert to polar
mag(xt) = 147.47, angle(xt) = 112.11 deg
So the answer is the magnitude of the result, 147.47 km.
If you need more detailed info on converting vector to polar or vice versa, visit the ref. page.