Velocity of rain in different reference frames

  • Thread starter dink87522
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  • #1
dink87522
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I'm having some trouble with the below vector question. I've got an answer although I'm not sure if it is correct or not.

Homework Statement



A car travels due east with a speed of 40 kph. Rain drops are falling at a constant speed vertically with respect to Earth. The traces of the rain on the side windows of the car make an angle of 50 degrees with the vertical. Find the velocity of the rain in the following reference frames: a) the car b) the Earth


Homework Equations



x = VxCosTheta
Y = VySinTheta
Trig

The Attempt at a Solution



X = (40)/(3.6 * tan 50) = 17
Y = (40)/(3.6 * sin 50) = 14.5

Would it be a) 17 m/s b) 14.5m/s ?
 

Answers and Replies

  • #2
N-Gin
56
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Just look at the vector relationships between two reference frames.

Let [tex]\vec{v}[/tex] be velocity of the rain in the reference frame of the Earth, [tex]\vec{v'}[/tex] velocity of the rain in the reference frame of the car and [tex]\vec{V}[/tex] be velocity of the car.

Then we have this (pretty much obvious) relation,

[tex]\vec{v}=\vec{V}+\vec{v'} \Rightarrow \vec{v'}=\vec{v}-\vec{V}[/tex].

Now, if you draw vector diagram, you can clearly see direction of the rain in both references. Let [tex]\alpha[/tex] be the given angle. Then, from geometry of the problem, we get

[tex]\tan{\alpha}=\frac{v}{V} \Rightarrow v=V \tan{\alpha}[/tex].

From there you can get [tex]v'[/tex] using Pythagorean theorem.
 

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