In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west.

Question

In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west. This average velocity results, because she hikes for 5.89 km with an average velocity of 2.18 m/s due west, turns around, and hikes with an average velocity of 0.555 m/s due east. How far east did she walk (in kilometers)?

 

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General Physics PRemastered 2 years 1 Answer 711 views 0

Answer ( 1 )

  1. You need to find the distance east, you know the velocity east so you just need to find the time spent traveling east, using the average velocity formula

    take west as positive direction, east as negative

    V = 1.27m/s
    v1 = 2.18m/s
    t1 = 5890/2.18 = 2702s …..time traveling west
    v2 = -0.555m/s
    t2 = ?………………………….time traveling east

    d1 = v1t1 = 5890m ……….displacement west
    d2 = v2t2 = -0.555t2………displacement east

    final displacement is
    D = d1 + d2 = 5890 – 0.555t2
    the total time of travel is
    T = t1 + t2 = 2702 + t2
    the average velocity is
    V = 1.27 = D/T= (5890 – 0.555t2)/ (2702 + t2)
    gives
    1.27(2702 + t2) = (5890 – 0.555t2)
    expand and rearrange for t2
    t2 = (5890 – 1.27*2702) / (1.27 + 0.555) = 1347s
    is the time spent traveling east

    so the distance traveled east is
    |d2| = |v2t2| = 0.555*1347 = 748m = 0.748km

    Best answer

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