Finding the Speed of Rain in Relative Motion

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Homework Help Overview

The problem involves relative motion, specifically analyzing the speed of rain as observed by a man walking at different speeds. The original poster seeks assistance in understanding how to approach the problem based on the given conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of vector triangles to represent the velocities involved. There are attempts to clarify the relationships between the velocities of the man and the rain, with some questioning the interpretation of the problem setup.

Discussion Status

Participants are actively engaging in the discussion, with some providing guidance on drawing vector diagrams and clarifying the relationships between the different velocities. There is an ongoing exploration of the correct interpretation of the problem, with no explicit consensus reached yet.

Contextual Notes

Some participants note potential misinterpretations of the problem, particularly regarding the direction of the rain's velocity and the setup of the vector triangles. The original poster expresses confusion about the teacher's explanation of relative motion.

kalupahana
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Relative motion prob, please help

Homework Statement


A man walk at a rate of 2 km/h; A rain appears to fall vertically. when he doubled his speed it appears to fall at 30o vertical. Find the real speed of the rain.


Homework Equations





The Attempt at a Solution


Help me to do this question. I cannot understand what my teacher said about relative motion. Please explain me how find the answer for this question.
 
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hi kalupahana! :wink:

let VMG be the velocity of the Man relative to the Ground,
VRG be the velocity of the Rain relative to the Ground,
and VRM be the velocity of the Rain relative to the Man

then draw a vector triangle MRG for each of the two situations: the length of the RG side will be the same in both triangles, and the length of the MG side will be 2 in one triangle and 4 in the other triangle :smile:

(see the pf library entry on vector triangle for some more details)
 


tiny-tim said:
hi kalupahana! :wink:

let VMG be the velocity of the Man relative to the Ground,
VRG be the velocity of the Rain relative to the Ground,
and VRM be the velocity of the Rain relative to the Man

then draw a vector triangle MRG for each of the two situations: the length of the RG side will be the same in both triangles, and the length of the MG side will be 2 in one triangle and 4 in the other triangle :smile:

(see the pf library entry on vector triangle for some more details)

In both instance VRM is constant.
then sin60 = 2/x

x = 4/√3 = 2.3 km/h

is this right
 
kalupahana said:
In both instance VRM is constant.

No, VRG is constant (it's the same Rain, so its velocity relative to the Ground is the same).

Try again … draw the two triangles, with a shared side …

what does it look like? :smile:
 


tiny-tim said:
No, VRG is constant (it's the same Rain, so its velocity relative to the Ground is the same).

Try again … draw the two triangles, with a shared side …

what does it look like? :smile:

I got triangle like this

then tan60 = VRG /4
4√3 = VRG
 

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hi kalupahana! :smile:

(just got up :zzz: …)

That's the right idea, but you've misinterpreted the question.

Your diagram show the rain falling vertically, but it doesn't.

Also, your arrows are going the wrong way round each vector triangle …

they have to match up so that VRG = VRM + VMG

(or VRM = VRG + VGM = VRG - VMG).

Try again. :smile:
 

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