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Finding average velocity

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Starting from the front door of your ranch house, you walk 55.0m due east to your windmill, and then you turn around and slowly walk 40.0m west to a bench where you sit and watch the sunrise. It takes you 25.0s to walk from your house to the windmill and then 47.0s to walk from the windmill to the bench.

    For the entire trip from your front door to the bench, what is your average velocity?

    2. Relevant equations

    [itex]\upsilon[/itex]av-x=x2-x1/t2-t1=Δx/Δt

    3. The attempt at a solution
    40.0m-55.0m/47.0s+25.0s = -0.208m/s

    So I got the answer to this problem by adding the time instead of subtracting it. The program I'm using shows this answer is correct. Though, I'm wondering why adding the time together instead of subtracting it works instead of the other way around. Can someone help explain this to me? Thanks.
     
  2. jcsd
  3. Aug 31, 2014 #2

    nrqed

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    ##t_1## and ##t_2## represent the time (on a clock) at the initial and at the final positions. Their difference ## t_2 - t_1 = \Delta t## is the time it took for the entire motion.
    This information is not given in the question. What they give is the time it takes for the two paths, so what they are giving is actually ##\Delta t_{first \, part} ## and ## \Delta t_{second \, part} ##. Since these are time intervals, you must add them to get the total time for the whole motion.
     
  4. Aug 31, 2014 #3
    t2 is the time you are at the bench.
    t1 is the time you start at ( call it 0).

    So Δt=t2-t1=(47.0s+25.0s)-0s
     
  5. Aug 31, 2014 #4

    PeroK

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    You are misunderstanding this equation:

    x1 is the starting position; and x2 the final position.

    t1 is the time at the start; and t2 is the time at the end.

    It might help to think of this equation as:

    [tex]v_{average} = (x_{final}-x_{start})/(t_{final}-t_{start})[/tex]

    Also, for many problems, not all so be careful, you can choose t_start = 0. That's when you start the stopwatch. t_end is when you stop the stopwatch. And ##\Delta t## is what the stopwatch reads.
     
  6. Aug 31, 2014 #5

    ehild

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    The route consist of two stages. You can consider the initial position at the house as xi=0 and the initial time Ti=0. First, you walk due East - your displacement is Δx1= 55 m, and the time is Δt1=25 s. Your position is x1=55 m with respect to the house. Then you turn back moving to west: your displacement is negative, Δx2=-40 m and it took the time Δt2=47 s. Your position is x2= Δx1+Δx2=55-40=15 m measured to East from the house. Your total displacement is Δx=15 m to East, and you walked for 47+25=71 s. The average velocity is Vav=Δx/Δt = (15-0)/71 , and it points to East.

    ehild
     
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