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A person moving east finds wind blowing south. If he moves with thrice his velocity without changing direction, then he finds the wind blowing in the south-west direction. Find the magnitude of actual speed of the wind if the velocity of the person is V.

Let W be the velocity of wind, and assume coordinate axes according to convention.

For the first part:

The velocity of the guy is Vi and the relative velocity of wind to the person is -aj, where ‘a’ is the winds relative velocity to the person.

So $$-aj=W-Vi$$ $$W=Vi-aj$$

For the second part:

The velocity of the person is 3Vi, and the velocity of wind (relative) is $-\frac{a}{\sqrt 2}i-\frac{a}{\sqrt 2}j$ (since it’s southwest, so it makes an angle of 45). Again, using the relative velocity equation $$W=3Vi-\frac{a}{\sqrt 2}i-\frac{a}{\sqrt 2}j$$ Now this is very most of you have realized that I am wrong, because I when I equate the above two expressions, by result is is very amusing. I have spent hours in this ‘simple’ question but I don’t want to waste anymore time.

Can you please help me with it?

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Aditya
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    Why do you use the same relative wind speed $a$ in both cases? All you know is the relative direction of the wind in both cases, not the relative speed. The relative speeds would be different. Then you have two variables, and need to get two equations to solve them - which you can do because you have to equate the coefficients of $i$ and $j$. – Jaap Scherphuis Sep 04 '19 at 13:37

1 Answers1

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You could examine the south component (or $-aj)$ of the wind and set up the following relationship

$$W_s^2=W^2-V^2=(2V)^2$$

So, you get

$$W=\sqrt{5}V$$

Equivalently, in your setup, you would have

$$W=Vi-aj$$ $$W=3Vi-ai-aj$$

which allows you to get $a=2V$ and in turn

$$W=\sqrt{V^2+(2V)^2}=\sqrt{5}V$$

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