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Homework Help: Velocity question

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    A person runs around an equilateral triangle. Each side is 53.3 meters. The person runs completely around the triangle once and 2/3's of the way around the triangle, and the average speed for this trip is 10 m/s. What is the magnitude of the amount of the average velocity for this trip? How do you do this??

    2. Relevant equations
    velocity = distance over time.

    3. The attempt at a solution
    the guy ran 53.3 meters x 5 which = a total of 266.5 meters. I divided that by 10 and get 26.65. But thats not the answer. Whats the answer and how do you do it?
  2. jcsd
  3. Sep 7, 2013 #2


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    You divided distance by speed, what did you got? You need to find the magnitude of average velocity.

    Velocity is a vector, and it does change along the triangle. What would be the average velocity if the person run once along the triangle?

  4. Sep 7, 2013 #3
    that does not help me at all
  5. Sep 7, 2013 #4


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    I try to guide you, but you need to show some effort. Check your notes or book. What is velocity?

  6. Sep 7, 2013 #5
    Check Again! :wink:
    Your equation for velocity is wrong. That gives speed.
    EDIT:Whoops! crossed posts with ehild.
  7. Sep 8, 2013 #6
    Wait i think i get where you are taking me. the displacement is 53.5 meters. He ran 5 sides of the triangle in 10 seconds. 10 divided by 5 = 2. The answer is 2 because it takes him 2 seconds to get from where he started to where he ended? I know the answer is 2, but is that why. Thank echild
  8. Sep 8, 2013 #7
    mmm....You are confusing distance with sides now.
  9. Sep 8, 2013 #8
    well he ran the whole triangle once which is 3 sides and then he ran 2 more sides which makes 5 total. his total displacement doesnt matter though am i correct? His displacement is just 1 side. and he runs each side in 2 seconds because 10 seconds divided by 5 sides is 2. i know the answer is 2 but is my reasoning correct?
  10. Sep 8, 2013 #9
    What are you supposed to find time or avg. velocity?
  11. Sep 8, 2013 #10
    Nope, you are not, not when you are supposed to find velocity. You have either confused definition of displacement or that of velocity.

    What is one side?
  12. Sep 8, 2013 #11
    "What is the magnitude of the amount of the average velocity for this trip?" I have to find avg velocity. I have the answer on my study guide, the answer is 2. I just want to know how to get to 2 so i can perform it on the test
  13. Sep 8, 2013 #12


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    Yes, the average velocity has magnitude 2m/s. But it did not take 2 seconds to reach to the final position. If L is the length of one side of the triangle, the person covers one side in t=L/v(speed) =53.5/10=5.35 s.
    Running along five sides of length L each, the person is one side away from its initial position. The magnitude of the displacement is L and the distance covered (the length of the whole path) is 5L. Velocity is displacement over time, velocity=L/t; speed is distance covered over time speed=5L/t.

    Last edited: Sep 8, 2013
  14. Sep 8, 2013 #13
    In your first post, you calculated the number 26.65. This is actually the total number of seconds it took for him to run the 5 sides of the triangle (rather than a velocity or speed). In another posting, you determined that his total displacement was 53.3 meters. Average velocity is defined as the total displacement divided by the total time.
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