Train in the Rain

Date: 11/21/96 at 21:41:42
From: EL-EL & JOSARGE &J.H.
Subject: Train window and raindrops

Dr. Math,

Help me with this problem:
  
During a rainstorm, the path of the raindrops appears to form an 
angle of 30 degrees with the vertical and to be directed to the left 
when observed from the side window of a train moving at 10 mph.  A 
short time later, after the speed of the train has increased to 15 
mph, the angle between the vertical and the path of the drops appears 
to be 45 degrees.  If the train were stopped, at what angle and with 
what velocity would an observer see the drops fall?
      
As the speed of the train increases, the angle between the vertical 
and the path of the drops becomes equal to 60 degrees.  Determine the 
speed of the train at that time. 
      
stdt.

Date: 11/22/96 at 12:59:28
From: Doctor Anthony
Subject: Re: Train window and raindrops

For relative velocity problems it is a good idea to bring the observer 
to rest by imposing on the whole system a velocity equal and opposite 
to his speed. Then the other moving body (in this case the rain) will 
move with a combination of its true velocity and that imposed (i.e., 
the equal and opposite velocity of the observer).

You can construct a sketch diagram to show what is happening by 
drawing a horizontal line of 10 units. At its left hand end draw a 
line at 60 degrees upwards to the right. Now extend the horizontal 
line 5 more units to the left, and draw another line upwards at 45 
degrees to the right from this new endpoint. The point at which these 
two lines meet is the one we need in order to be able to complete the 
velocity diagram. Starting at the left hand end and moving clockwise 
round this figure, label the points A, B, C, and D. Thus CD = 10 
units, CA = 15 units, and BC = actual speed of the rain.

We can calculate AB using the sine formula in triangle ABD because we 
have the required two angles and a side:

AB = 5 x sin(120)/sin(15) = 16.73

Then BC^2 = 16.73^2 + 15^2 - 2 x 16.73 x 15 x cos(45)

          = 149.9996   

    so BC = 12.247

Angle BCA is found from sin(C)/16.73 = sin(45)/12.247

                              sin(C) = 0.965942 

                                   C = 75 degrees

So the angle with the vertical is 15 degrees to the right.

The speed of the rain is 12.247 mph slanting at 15 degrees in the 
direction of motion of the train.

The last part of the problem is easily done using the sine formula.  
If v is the velocity of the train, then:

  v/sin(75) = 12.247/sin(30)   so  v = 23.659 mph

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/