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March 25, 2017

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A jogger accelerates from rest to 5.64 m/s in 3.36 s. A car accelerates from 24.6 to 39.8 m/s also in 3.36 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 3.36 s?

  • physics - ,

    a. a = (V-Vo)/t = (5.64-0)/3.36=1.68 m/s^2.
    d1 = 0.5a*t^2 = 0.5*1.68*(3.36)^2=9.48 m.

    b. a = (39.8-24.6)/3.36 = 4.52 m/s^2.
    d2 = 0.5*4.52*(3.36)^2 = 25.5 m.

    c. d2 d1 = 25.5 9.48 = 16.0 m.

  • physics - ,

    c. d2-d1 = 25.5-9.48 = 16.0 m.

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