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Average velocity question

  1. Aug 14, 2011 #1
    hey guys im very new to physics and its just not making sense at the moment and i know this is a very simple question but i just dont get how to get the velocity !! ive worked out the average speed but i have no idea what the next step is to get the velocity, any hints would be very much appreciated cheers =)

    1. 2.A straight track is 1600 m in length. A runner begins at the starting line, runs due east for the full length of the track, turns around, and runs halfway back. The time for this run is five minutes.
    What is the runner’s average velocity
    and what is his average speed?




    2. v=x/t



    3. what ive worked out so far is distance is 2400m and 5 minutes is 300s
    average speed = 2400m / 300s
    = 8.0 m/s
     
  2. jcsd
  3. Aug 14, 2011 #2

    PeterO

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    That's good for the average speed - since speed is distance / time.

    How do we calculate Velocity?
    hint: velocity is a vector.
     
  4. Aug 14, 2011 #3
    You got speed right, speed is distance over time.
    Velocity is actually displacement over time.
     
  5. Aug 14, 2011 #4
    ok so ive worked out that average velocity = 800m/300s which gives us 2.66 east (rounded up to 2.7) but i dont totally understand why ive used 800m, how is 800m the displacement, i thought the displacement would have been the 1600m in the east direction not 800 back to the west ?
     
  6. Aug 14, 2011 #5

    PeterO

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    ?????

    You gave the velocity as 2.7 m/s, East [correct]

    What do you mean "i thought the displacement would have been the 1600m in the east direction not 800 back to the west ?"

    800m East is what you get when you go 1600m East and THEN 800 m back to the West from there.
    Moving 800m to the west will not get you West of your original starting position???? [because you first went 1600m to the east]

    EDIT: Perhaps it would have been clearer if the trip had only included 700m West, so that displacement was 900 East and you would not have been distracted by the numerical values being the same.
     
    Last edited: Aug 14, 2011
  7. Aug 14, 2011 #6
    ohhhhh that makes total sense =)

    the difference between the path of the initial and final position being the definition of displacement

    the displacement then is actually 800m east from the initial position (1600 - 800, (800 being halfway back of 1600) as you stated)

    thanks heaps =)
     
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