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Rain drops relative velocity

  1. Oct 16, 2009 #1
    1. The problem statement, all variables and given/known data

    While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38 degrees with the vertical. While driving back home moments later at the same speed but in the opposite direction, you see that the rain is falling straight down.

    From these observations, determine the speed of the raindrops relative to the ground.

    From these observations, determine the angle of the raindrops relative to the ground.

    2. Relevant equations

    None given


    3. The attempt at a solution


    I believe I am misunderstanding the wording of the question. I thought that since the rain drops are falling vertical while you are driving south at 25 m/s, it must mean that the rain drops are traveling 25 m/s in the direction that the car is traveling. I am at a loss of where to start. I know that 25 m/s is the wrong answer to the first question. Any ideas?
     
  2. jcsd
  3. Oct 16, 2009 #2

    Andrew Mason

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    Draw a vector diagram.

    Draw the vector for the velocity of the rain drop relative to the ground. From the tail of that vector draw the car velocity going north. Draw the difference vector, which is the velocity of the rain drop relative to the car. You know the direction of this vector: 38 degrees from vertical.

    Next, from the tail of the vector for the velocity of the rain drop relative to the ground again draw the car velocity going south. Draw the resultant difference vector, which is the velocity of the rain drop relative to the car. You know the direction of this vector: 0 degrees from vertical.

    Now you know two angles and a side of this triangle. Can you compute the other sides? Can you calculate now the length and direction of the first vector you drew?

    You might want to start with the difference vector and add the car velocity to get the velocity of the rain relative to the road. Do whatever is easier.

    The reason 25 m/s is the wrong answer is because you are not factoring in the vertical speed.

    AM
     
    Last edited: Oct 16, 2009
  4. Oct 16, 2009 #3
    Thank you for your response.

    Everything you have said makes sense to me except for one part. I drew the first vector (the velocity of the rain drop relative to the ground) from the origin towards the positive x-axis with an angle of 0 going counterclockwise from the origin. Is this right? Next, as you said, I drew a vector going from the tail of the vector Vrain (the velocity of the rain drop to the ground) heading directly north (Vcar). Next I drew the difference vector Vdifference. I drew this vector as going from the head of the Vcar to the head of Vrain. I think I have gone wrong somewhere here.

    To continue, I labeled Vcar as 25, the angle between Vcar and Vdifference I labeled as 38 degrees. I solved the sides of the triangle as the magnitude of Vdifference = 31.73 and the magnitude of Vrain = 19.5.

    I would like to stop here because I believe I am going about this wrong. Could someone either dispute or confirm my approach?
     
  5. Oct 17, 2009 #4

    Andrew Mason

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    You know the direction and magnitude of the car:ground velocities. But you know only the directions of the rain:car velocities. You have to determine their magnitude.

    The velocity of the rain relative to the car is the difference between the velocity of the rain relative to the ground and the velocity of the car relative to the ground. In other words, the velocity of car:ground + velocity of rain:car = velocity of rain:ground.

    You might find it easier to draw the velocity of the car:ground + velocity rain:car to provide to provide two triangles which have the rain:ground velocity as a common vector (the sum of each car:ground and rain:car vector = the same rain:ground vector).

    The base of the outer triangle has length 2 x velocity car:ground and you know the three angles (90, 38 and 52). Solve for the length of the other two outer sides (using triangle geometry). Then work out the length of the middle line (the rain:ground velocity vector).

    AM
     
  6. Oct 17, 2009 #5
    I am still not understanding this. I have some experience with vector arithmetic. I believe that I understand the concept. Most importantly I understand:

    "The velocity of the rain relative to the car is the difference between the velocity of the rain relative to the ground and the velocity of the car relative to the ground. In other words, the velocity of car:ground + velocity of rain:car = velocity of rain:ground."

    What I don't understand is the most fundamental part of this. I don't understand how to draw this situation.

    "You might find it easier to draw the velocity of the car:ground + velocity rain:car to provide to provide two triangles which have the rain:ground velocity as a common vector (the sum of each car:ground and rain:car vector = the same rain:ground vector)."

    This is what I do when I attempt to draw the above suggestion. I first draw a vector of magnitude 25 at an angle of 90 degrees. From the head of this vector I draw another vector extending at a 38 degree angle of unknown magnitude. From the tail of the first vector I drew, I draw a vector that connects to the head of the second vector I drew.

    I have no understanding of how this could be correct. Mostly because the magnitude could be anything.

    This problem frustrates me. It frustrates me because I know I can easily solve triangles and I have not had much trouble understanding vectors in general. I am having problems understanding the wording of this problem.

    How do you interpret "38 degrees with the vertical"? Should I be viewing this situation in the zy-plane or the xy-plane?

    Thank you Andrew, for all of your help. I'm sorry I'm not understanding this easier.
     
  7. Oct 18, 2009 #6

    Andrew Mason

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    Draw a vector 25 units long from the origin to the left. Call that vector cgs. From the head of vector draw a vector pointing straight down of unknown length. Just make it arbitrarily long. Call that vector rcs. Go back to the origin. Draw a vector 25 units long from the origin to the right. Call that vector cgn. From the head of that vector draw a vector pointing down at 38 degrees from the vertical toward vector rcs, until it intersects with rcs. Call that rcn.

    Now, you know that cgn + rcn = rg. You also know that cgs + rcs = rg. So: cgn + rcn = cgs + rcs.

    You just have to work out the length and direction of rg using simple triangle geometry.


    AM
     
  8. Oct 23, 2009 #7
    Ahh, I see now.

    Thank you once again for your help. That clears things up for me.
     
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