# man walking in rain:relative velocity [closed]

A man walking in rain at a speed of 3kmph find rain to be falling vertically. When he increases his speed to 6 kmph, he finds rain meeting him at an angle of 45 deg to the vertical. What is the speed of the rain??

My approach:

$\overline{V_{\operatorname{rain}}} - \overline{V_{\operatorname{man}}} = \overline{V_{\operatorname{rain relative to man}}}$

$\Longrightarrow \overline{V_{\operatorname{rain}}} = \overline{V_{\operatorname{rain relative to man}}} + \overline{V_{\operatorname{man}}}$

Equating velocity of rain wrt ground in both cases, we get

$9+ x^2 = 36+ x^2 - 12x \frac{1}{\sqrt{2}}$ where x is velocity of rain wrt to man.

so x becomes $9\frac{\sqrt{2}}{4}$ but that means velocity of rain wrt ground is (according to the first case, where velocity of man is 3 kmph) :

$\sqrt{9 + {(9 \frac{\sqrt{2}}{4}})^2}$

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## closed as off-topic by Manishearth♦Jul 7 '13 at 9:41

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You may want to check out this video: Is it Better to Walk or Run in the Rain?. –  Māris Ozols May 2 '13 at 13:55
I don't see how that can help me. That vid only debates about the amount of water that soaks you, in relation to time. I think a solution using relative velocity concept is what the question demands. –  Saurabh Raje May 2 '13 at 14:00

The right hand side of your first equation involving $x$ is wrong. I will use $v_y$ to mean your $x$. Then you would have $9+v_y^2$ for the squared speed of the rain in the first case as you say. When he starts walking faster the squared speed of the rain is $(6-v_y)^2 + v_y^2$. The reason for the $6-v_y$, is that the man sees the angle of the rain as $45^\circ$ so the $x$ velocity must be equal to the $y$ velocity. And since the relative $x$ velocity was zero when he was moving more slowly, the relative $x$ velocity must be negative when he is moving more quickly.
The faster way to do the problem is just to take from the part where he is walking more slowly that the $x$ velocity of the rain is 3 kmph (since it equals the $x$ velocity of the guy). Then take from the second part that the magnitude of the relative $x$ velocity must be 3kmph, which must be equal to the magnitude of the $y$ velocity since the angle is $45^\circ$. So the $y$ velocity of the rain must be at 3 kmph. So the speed is $3 \sqrt{2}$ kmph. The other way works too though.
so doesnt that make it $3\sqrt{2}$?? ie the speed of rain wrt to ground is $3\sqrt{2}$. That is one the options. btw, thanks for the shorter method...i now feel like a fool! I ignored the 45 deg part...had figured out the x component of the rain velocity to be 3 kmph... –  Saurabh Raje May 2 '13 at 14:15
you're right I just solved for the $y$ velocity. I know edited the answer to find the speed of the rain. The speed of the rain $3 \sqrt{2}$ as you said. In order to explain why your equation is wrong I would need you to explain how you got it. Otherwise all I can say is that its wrong. The $(6-v_y)^2 + v_y^2$ is the squared speed of the rain in the frame of the man when he is moving faster. –  NowIGetToLearnWhatAHeadIs May 2 '13 at 14:20
$vector( a) + vector( b )= \sqrt{a^2 +b^2 + 2abcos\theta}$ –  Saurabh Raje May 2 '13 at 14:22
cos(135) = - 1/($\sqrt{2}$) –  Saurabh Raje May 2 '13 at 14:23