Finding magnitude and direction question

Question

You leave the airport in College Station and fly  27 km in a direction 34.0  south of east. You then fly 46 km due north. How far must you then fly to reach a private landing strip that is 32.0 km due west of the College Station airport? In what direction?

Here is my work

Rx = 27(cos(34)) - 32 = -9.6

Ry = -27(sin(34)) + 46 = 30

(-9.6)^2 + (30)^2 = c^2

c = 31.5

I'm not sure what I'm doing wrong. I keep getting 31.5 km.

Answer 1

Hey Karen,

I think your issue is with including the desired resultant vector into your calculations. The way you broke down the first 27km vector looks great, as does the way you added the 46km path north. But then, the question is asking basically what vector you would have to ADD to the existing flight vector (which is the sum of the 27km and 46km paths) in order to end up at a new vector that is just Dx = -32.0km west. So, if you draw this out, the plane is somewhere in the northeast, and it has to make a journey back south AND all the way over to 32km west.

1. Remove the -32 from your Rx expression.
2. Create a new equation involving both x and y components such that R + ?? = -32km West.

Good luck!

Answer 2

Consider the signs for your Rx calculation. You initially fly east, however the landing strip is due west. So you need to subtract -32km (the same as adding 32km) in your Rx calculation. Try:

Rx=27km*(cos(34 deg)) + 32km=54.38km