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I'm really lost with average velocity and displacement!

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Heather and Matthew walk with an average velocity of 0.98m/s eastward. If it takes them 34 min to walk to the store, what is their displacement?

    2. Relevant equations
    V avg = Δxt (since I'm looking for the displacement I changed it to) Δx=V avgΔt

    3. The attempt at a solution
    Δx=VavgΔt=(0.98 m/s)(34min)=33.32m East

    When I looked at the back of the textbook it said 2.00km East, how did they get that number?
  2. jcsd
  3. Sep 20, 2010 #2
    You should check your units. Make sure everything is canceling out correctly.
  4. Sep 20, 2010 #3
    I don't get that. Sorry I just started physics. I was following an example from the textbook, and I understood that. But, the questions aren't making any since. I'm starting to pull my hair trying to get the right answer.
  5. Sep 20, 2010 #4


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    You have the right equation. As localrob has noted, check your units for the time taken. You have the time in minutes, but you should denote the time in units of ______?____.
  6. Sep 20, 2010 #5
    In units of seconds right. So would I have to covert 34 minutes to seconds?
  7. Sep 20, 2010 #6


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    Sure....(m/sec)(sec) = m
  8. Sep 20, 2010 #7
    Thanks I got it!

    Δx=(0.98m/s)(34 min * 60 sec/min)
    which gives me 1,999.2/1000 = 1.9 or 2.0km
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