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Need help with relative velocity problem

  1. Feb 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Amy is traveling at 60mph on the highway, due North. The on-ramp nearby starts out going West, and then curves around so that eventually it is also North, parallel to the highway.
    A) Bob appears, traveling 40mph going West (on the onramp). What is his speed relative to Amy?
    B) In the next moment, he is heading due Northwest, now going 50mph. What is his speed relative to Amy now?
    C) Finally, he is in the Northward position, and is accelerating to 55mph. What is his speed relative to Amy now?

    2. Relevant equations
    I think this involves some basic trig to calculate the vector components / actual vectors.

    3. The attempt at a solution
    So I am a little fuzzy on the whole relative velocity when objects are moving towards eachother at angles. For the last part, (C), I believe it would be 5mph, because relative to her speed, he is going 5mph slower that she is.

    I attempted part (A) by setting up a right triangle where the Y-Component is 60mph, the X-Component is 40mph, and solved for the resultant vector, which yielded √602 + 402 = 72.1 mph. I am not entirely sure if this is even the correct approach however.

    For part (B), I am a little confused about how exactly the diagram should look like. All I know is that Northwest would be 45°, but I don't how the triangle itself would look.
  2. jcsd
  3. Feb 29, 2016 #2
  4. Feb 29, 2016 #3
    Funny, I was actually looking at that page yesterday when I was attempting it. It helped me understand how I might create the diagram, but I still feel don't feel confident about it. Here is what I think the diagram looks like:
    But again, I'm not sure.
  5. Feb 29, 2016 #4
    That is right :D, Now you have two vectors.
    Two get the relative velocity you need to add these vectors up. If a vector has an angle then it has two components vertical and horizontal.
    Amy's velocity vector doesn't angle so it only has y component.

    So lets just break them into two components,
    Vx = 50 cos Theta
    Vy = 50 sin theta

    What about amy? can you figure it out?
    Now if draw them up and add them as usual if they are in the same direction then you have to add them if they are in opposite direction you have to subtract them.
    Once you finish that, You will end up with 1 vector in the y direction and 1 vector in the x direction. Now you get resultant vector through pythagoras.
  6. Feb 29, 2016 #5
    I'm not sure if I did it right, here's what I did:

    Vx = 50cos45° = 35.35 i
    Vy = 50sin45° = 35.35 j

    Vx = 0 i
    Vy = 60 j

    (35.35 i + 0 i) + (35.35 j + 60 j ) ==> 35.35 i + 95.35 j ==> √35.352 + 95.352 = 101.69mph
  7. Feb 29, 2016 #6
    Well done, But just a question.. What does I and J represent?
  8. Feb 29, 2016 #7
    i and j are just a fancy way of saying x-direction and y-direction. So i = x-direction and j = y-direction. I think k = z-direction, but I haven't learned that yet. Btw, thanks for the help, I appreciate it a lot!
  9. Feb 29, 2016 #8
    Oh, No problem :D
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