The problem is as follows:

A person is running following a constant speed of $4.5\,\frac{m}{s}$ over a flat (horizontal) track on a rainy day. The water droplets fall vertically with a measured speed of $6\,\frac{m}{s}$. Find the speed in $\frac{m}{s}$ of the water droplet as seen by the person running. Find the angle to the vertical should his umbrella be inclined to get wet the less possible. (You may use the relationship of $37^{\circ}-53^{\circ}-90^{\circ}$ for the $3-4-5$ right triangle ).

The alternatives given on my book are as follows:

$\begin{array}{ll} 1.7.5\,\frac{m}{s};\,37^{\circ}\\ 2.7.5\,\frac{m}{s};\,53^{\circ}\\ 3.10\,\frac{m}{s};\,37^{\circ}\\ 4.10\,\frac{m}{s};\,53^{\circ}\\ 5.12.5\,\frac{m}{s};\,37^{\circ}\\ \end{array}$

On this problem I'm really very lost. What sort of equation should I use to get the vectors or the angles and most importantly to get the relative speed which is what is being asked.

I assume that to find the relative speed can be obtaining by subtracting the speed from which the water droplet is falling to what the person is running. But He is in this case running horizontally. How can I subtract these?

Can somebody give me some help here?

Supposedly the answer is the first option or $1$. But I have no idea how to get there.

  • $\begingroup$ The $3-4-5$ triangle is a hint as to the nature of the result. He is traveling horizontally, the droplets travel vertically, the resulting relative speed must combine these quantities in some manner. Since the two lines of travel are perpendicular, it makes sense to think about the hypotenuse between them... $\endgroup$
    – abiessu
    Nov 4 '19 at 3:43
  • $\begingroup$ @abiessu Perhaps can you develop an answer so I could understand what you mean by combining the speeds?. Can you show me the equation should I use?. Had I not been given the alternatives, what should I have done with this question? $\endgroup$ Nov 4 '19 at 4:27

Since the rain is vertical, the runner runs into falling droplets. To him is like the droplets fall slanted towards him.

You have a right triangle formed with the opposite of the speed of the runner and the speed of the droplet as legs abs the resulting speed of the droplet as the hypothenuse.

The first answer is correct.

  • $\begingroup$ I'm sorry. But i'm still stuck. Perhaps can you uncompress your answer a little bit. Because I feel it is too condensed with words. Can you develop into equations so I can understand what's going on?. I try to get the picture in my head, so that what is happening is that a right triangle is formed?. But I don't know what should I use or how should I put the vectors there. $\endgroup$ Nov 4 '19 at 4:26
  • $\begingroup$ water droplet speed as seen from a stationary point on earth is $\vect{v_d}=(0,-6)$. the speed of the runner as seen from the same point is $\vect{v_r}=(4.5,0)$. however, from the perspective of the runner, the water droplet both falls down with the speed $\vect{v_d}=(0,-6)$ and it moves towards him with the horizontal speed $\vect{v_h}=(-4.5,0)$. wherefore, for the runner the water moves with a resultant speed $\vect{v_r}=\vect{v_d}+\vect{v_h}=(0,-6)+(-4.5,0)=(-4.5,-6)$. this vector has a magnidue of 7.5 and a direction of $37^0$ West from vertical. $\endgroup$
    – WindSoul
    Nov 4 '19 at 6:49
  • $\begingroup$ Should I conclude that relative speed will always be a sum as in this case? It gives me impresion. But how to avoid getting confused. Lets say im in a bus and I see through the window another bus comes in the opposite direction im moving. Will the real speed of the bus be the difference of the speed from which i see the bus minus the speed of the bus where i am at? But if the answer is yes? How come in this case it doesnt seem to be a sum? How about a scerario where i see another bus which passes by in my same direction but faster. Should the real speed of that bus be the sum or difference? $\endgroup$ Nov 4 '19 at 17:14
  • $\begingroup$ By the way I believe your previous comment should be upgraded to be part of your answer as it is aiding me much better than the original answer. $\endgroup$ Nov 4 '19 at 17:16
  • 1
    $\begingroup$ Yes. Is called composition of vectors, to avoid the confusion around addition or subtraction. $\endgroup$
    – WindSoul
    Nov 5 '19 at 4:23

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