# Relative motion of airplane in the wind [closed]

I have a relative motion problem in which I cannot get my answer to match the book answer. The question is:

An airplane has to travel $189\, \mathrm{km}$ due east to point $B$ from point $A$. It can fly at $100\, \mathrm{kph}$ in still air. There is a wind at $60\, \mathrm{kph}$ blowing from the south. Find the time taken for the trip.

This is what I have done: $$Vw=60y$$ $$Vpw=100 \cos(\alpha)x -100 \sin(\alpha)y$$ $$\therefore Vp=Vpw+Vw$$ $$Vp=100 \cos(\alpha)x - 100 \sin(\alpha)+60y$$ The y velocity for the plane is 0 as it is traveling east, so I get the following: $$\sin(\alpha)=\frac{6}{10}$$ $$\therefore \cos(\alpha)=\frac{4}{5}$$ $$\therefore Vp=100(\frac{4}{5})x=80$$ Therefor Time taken should be $\frac{189}{80}=2.3625\, \mathrm{hours}$ The answer in my book says the answer should be 2 hours.

Could someone show me where I am going wrong, thanks.

## closed as off-topic by David Z♦Nov 22 '13 at 1:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• Your answer is correct, the book can be wrong sometimes – Pranav Hosangadi Nov 21 '13 at 23:53
• @DavidZ, Voting to reopen. The OP has shown significant effort, has arrived at the correct answer, and is perplexed because the textbook has a typo. If this site objects to providing help in these cases then I shall endeavour to fix that. – Alfred Centauri Nov 22 '13 at 1:55
• @Alfred there's no specific conceptual question here though. The OP is just asking for someone to check their work. If you think questions of that nature should be allowed, I'd encourage you to voice your concerns on Physics Meta on one of the discussions of the homework policy. – David Z Nov 22 '13 at 2:08
• It's a 3-4-5 triangle. The southeast hypotenuse is 100, the south side is 60, so the east side is 80 kph. 189/80 = your answer. – Mike Dunlavey Nov 22 '13 at 13:46

There's a more straightforward calculation.

In order to travel eastward, the plane's velocity must have a southward component of 60kph to cancel the wind from the south.

Since the plane's speed is 100kph, we have the eastward component (in kph) is just:

$$v_E = \sqrt{100^2 - 60^2} = 80$$

Thus, you are correct; the time required to travel 189km eastward (in hours) is:

$$t = \dfrac{189}{80} = 2.36$$

• I used that method as well but the wrong answer in the book threw me off. Thanks – user Nov 22 '13 at 0:27