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One question

  1. Sep 30, 2005 #1
    New to physics, and I'm a little confused on one problem on my homework:

    1) A runner runs 5.0km along the course east, then turns around and runs 5.0km west along the same path. She returns to the starting point in 40 min.

    What is her average speed and velocity?

    The speed was easy to get; .25km/h

    I'm not sure if I'm doing this right, but I keep ending up with something bizzare like 0/-40 for the average velocity.

    I mean, the equation IS d sub 1 - d sub 2
    t sub 1 - t sub 2

    Sorry if its a dumb question, but am I doing anything wrong?

    - A confused High school student
  2. jcsd
  3. Sep 30, 2005 #2


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    It's not a dumb question at all. It goes to the heart of the difference between distance and displacement. You get the average speed by doing exactly what you did - dividing the distance travelled by the time it took.

    You get the average velocity by dividing the displacement by the time. Displacement is a vector that runs from where you start to where you finish. What is her total displacement in this problem? Average velocity tells you the velocity she would have to have had to go from where she started to where she stopped without changing speed or direction. Given her displacement, what would that velocity be?
    Last edited: Sep 30, 2005
  4. Sep 30, 2005 #3
    So, then, her total displacement being 5km? So 5km divided by 40 min giving me .125km as the velocity?
  5. Sep 30, 2005 #4


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    Nope. Think again - if she runs 5km east, turns and runs 5km west, how far is she from her starting point when she's done?
  6. Sep 30, 2005 #5
    Zero, that is her final displacement.
  7. Sep 30, 2005 #6


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    And if her average velocity is displacement divided by time, what is her average velocity?
  8. Sep 30, 2005 #7
    That was my orginal answer, 0/40....so, the average velocity is zero...?

    That was easier than I thought! :uhh:
  9. Sep 30, 2005 #8


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    Yep. People say that a lot. Remember - since she ended up where she started, there's a sense in which she didn't go anywhere. The only way that she could have a constant velocity and end in the same place she started would be if she didn't move at all. Her average velocity would have to be zero.

    Does that make sense?
  10. Sep 30, 2005 #9
    Great! Thanks, these forums are going to be useful for this year.

    Just wanted to make sure I have some other things right.

    So, speed = distance/time, the amount of distance covered in a certain amount of time

    Velocity = Final displacement from starting point/time, shows how much you move in a certain direction

    One thing I wanted to make sure of; if there was a car going at 90km/h, and it passed a truck going 105km/h, then to the car's refrence point, the truck would be moving at 15km/h. Right?

    Thanks for the help! Now I can go to bed!
  11. Sep 30, 2005 #10


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    You need to make a distinction between speed (instantaneous speed), velocity (instantaneous velocity), average speed and average velocity.

    Speed is the thing the speedometer in your car measures. It's how fast you're going right now! It can change from one second to the next.

    Average speed is how far you travel (path length - the length of the path you followed) divided by time. If you could somehow get a time-weighted average of the readings of your speedometer, it would give you the average speed, but it's probably best right now just to think of it as the formula.

    Velocity is speed with a direction. Put a compass by your speedometer and you have a velocimeter. Velocity is a vector - direction matters. Travelling for one hour at 60 mph north and another hour at 60 mph south will give you a resulting average velocity of 0. They cancel out, because they're in opposite directions.

    Average velocity is the displacement over time.

    Although it is true that instantaneous speed is the magnitude of instantaneous velocity (i.e. instantaneous velocity minus the direction), the same is not true of average speed and average velocity. The time component in the averages screws up the simple relation.

    As for the car/truck thing, yes. Exactly.
  12. Sep 30, 2005 #11


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    Apparently not all that easy! If she ran .25 km/h for only 2/3 of an hour (40 min.) she would have run much less than 1 km, not 10!

    Is it possible you meant .25 km/minute??
  13. Sep 30, 2005 #12


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    Oops. Thanks, HallsofIvy - I should've seen that.
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