Physics

A young woman named Kathy Kool buys a sports car that can accelerate at the rate of 4.77 m/s2. She decides to test the car by dragging with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.04 s before Kathy. If Stan moves with a constant acceleration of 3.59 m/s2 and Kathy maintains an acceleration of 4.77 m/s2, find the time it takes Kathy to overtake Stan.

Calculate the distance she travels before she catches him.

Calculate the speed of Kathy's car at the instant she overtakes Stan.

Calculate the speed of Stan's car at the instant he is overtaken by Kathy.

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asked by Mathew
  1. a. d1 = d2,
    0.5a*t^2 = 0.5a(0.5t-1.04)^2,
    0.5*3,59t^2 = 0.5*4.77(t-1.04)^2,
    1.795t^2 = 2.385(t-1.04)^2,
    0.7526t^2 = t^2-2.08t+1.082,
    0.7526t^2 -t^2+2.08t =1.082,
    -0,2474t^2 +2.08t - 1.082 = 0,
    Use Quadratic formula and get:
    t = 7.85 s.

    b. d = 0.5*a*t^2,
    d = 0.5*4.77*(7.85)^2 = 147 m.

    c. V = 4.77*7.85 = 37.44 m/s.

    d. V = 3.59 * (7.85+1.04) = 3i.9 m/s.

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    posted by Henry

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