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1. Carlos jogs in a straight line at a constant speed of 1.5 m/s. He passes by Victoria, who, 10 seconds after Carlos had passed her, starts accelerating at a constant rate of 0.50 m/s^2

a) How much time after the passing of Carlos does it take Victoria to catch up to him?

b) What distance did they travel?

  • Physicss -

    Assuming that Victoria was standing still when Carlos passed her, we have the following equations, calling t=0 when he passes her:

    Carlos: s = 1.5t
    Victoria: s = .25(t-10)^2

    set them equal to find when they have traveled the same distance -- that is, when she passes him.

    1.5t = .25(t^2 - 20t + 100)
    1.5t = t^2/4 - 5t + 25
    t^2/4 - 6.5t + 25 = 0
    t^2 - 26t + 100 = 0
    t = 13 ± √69
    t = 4.7, 21.3

    Since Victoria does not start till t=10, the only answer is 21.3, or 11.3 seconds after she started running.

    Plug in times to get distances.

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