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Your answer for question a is listed above in units of meters/second, but the work shows that the units are actually kilometers/hour.

Hi clarkmg,

Thank you very much for noticing that. I've fixed the Quick Answer to reflect the video.

Mr. Dychko

Dear Mr. Dychoko,

For part (a), I understand how you used the cosine rule to find the unknown vector C. However, I tried to solve the problem similar to what you did in problem 42 and divided the wind velocity into X and Y components. I used to pythagorean theorem to find the resultant vector but was not able to get the right answer. Can you please tell me why it is wrong to use this method.

Thank You

Hi chaegyunkang, thanks for the question. The method you describe works fine, so your sleuthing should look at how you implemented it. Pay careful attention to whether you subtracted the y-component of the velocity of the air with respect to the ground from the velocity of the plane with respect to the air (instead of adding it). Knowing that $\cos(45) = \dfrac{1}{\sqrt{2}}$, here's what the work will look like in your calculator:
$\Big(688 - \dfrac{90}{\sqrt{2}} \Big)^2 + \Big(\dfrac{90}{\sqrt{2}}\Big)^2 = \textrm{ some number}$
$\sqrt{ \textrm{ some number}} = 627.5953...$ which is the correct magnitude, and I'll leave the angle for you.

Hope this helps,
Mr. Dychko

My mistake was adding the vectors. Thank you so much.

You're welcome! That's a common mistake, so make sure you don't sweat it.