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Homework Help: Calculate your average velocity from pillar to post

  1. Sep 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Starting from a pillar, you run a distance 200m east (the x-direction) at an average speed of 5.0m/s , and then run a distance 280m west at an average speed of 4.0m/s to a post.

    Calculate your average speed from pillar to post.

    Calculate your average velocity from pillar to post.

    I drew a rough sketch on how I would this would look like:
    http://img365.imageshack.us/img365/449/untitled1zs5.jpg [Broken]

    I understand that they are on the sime line, as if the runner is running forward and back,
    I drew it in two seperate lines for convienience.

    2. Relevant equations

    V = X2-X/T2-T1

    3. The attempt at a solution

    At first, I would assume that on they way back, the speed would be 4.0m/s and would remain that for the 80m portion gap between 200 and 280m.

    Then I decided, that was too easy (and plus I got it wrong), so I think the equation shown above comes into play. But I don't know where to go from there, can anyone stir me into the right direction.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 6, 2007 #2

    learningphysics

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    Average speed and average velocity are different.

    Average speed = (Total Distance)/(Total Time)

    Find the total distance... find the total time for the trip... divide them for the average speed...

    Average velocity = (Total Displacement)/(Total Time) (displacement is a vector)
     
  4. Sep 6, 2007 #3


    Thank you for your fast response.

    I understand that Average speed is total distance over total time, which leads the av. speed to be 4.4 m/s.

    For Average velocity, you mention that it's total displacement over total time.

    The displacement would have to be 80m, -80m since it's going 80m from where it started.
    Total time is where I am starting to have problems,

    I know that the 200m @ 5m/s has a total time of 40 seconds.
    I know that the 280m @ 4m/s has a total time of 70 seconds.

    Would total time be 30 seconds? or will be still be 70+40= 110s.

    Average Velocity = -80m/30s
    or
    Average velocity = -80m/110s?
     
  5. Sep 6, 2007 #4

    learningphysics

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    Average speed looks right. Total time is 110s. So average velocity would be -80/110 m/s. And since east is positive and west is negative, in your final answer I would use 0.727 m/s West.
     
  6. Sep 6, 2007 #5
    Thank you, I was thinking about .727, but thought to myself that the number might be too small for average velocity.

    I understand that -80m is the displacement, and time is 110 as you mentioned total time. Time cannot be negative, so the final result would have to be -.727 m/s.

    I thank you again for the time you took to help me out. :smile:
     
  7. Sep 6, 2007 #6

    learningphysics

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    No prob. be sure at the end to interpret the negative sign as being west... ie: writing the final answer as 0.727m/s West.
     
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