CHAPTER 2

  1. a. At t = 1.0s, v = 4m/s, a = 0, x = 2.0m + (4 x  1)m = 6.0m.
    b. At t = 3.0s, v = 2m/s, from the slope of the line a = -(2/1)m/s2 = -0.5m/s2,  x = 2.0m + [4 x 2 + 2 x 1 + 1/2(2 x 1)] = 13.0m.
  2. a. vx = v0x + ò0tadt = 0 + ò0t(10 - t)dt;
        0 = 0 + (10t - 1/2t2)|0t = 10t - 1/2t2, which gives t = 20 s.
    b. x = x0 + ò0tvdt = 0 + ò0t(10t - 1/2t2)dt = (5t2 - 1/6t3)|0t =20s = 666.7 m.
  3.  
  4. a. x - x0 = 110m - v0t = 110m - (20m/s)((0.5s) = 100m when apply brake.
    b. We find the acceleration from
        v2 = v02 + 2a(x - x0);
        0 = (20m/s)2 + 2a(110m -10m), which gives a = -2m/s2.
    c. From v = v0 + at, we find the time t = (v - v0)/a = (0-20m/s)/(-2m/s2) = 10s.
  5. a. We find the time for the first splash
         y = y0 + v0yt + 1/2gt2;
        50m = 0 + (20m/s)t + 1/2(9.8m/s2)t2, which give t1 = 1.75s
        We find the time for the second splash from
         y = y0 + v0yt + 1/2gt2;
        50m = 0 + (-20m/s)t + 1/2(9.8m/s2)t2, which gives t2 = 5.83s.
        The time elapse is Dt = t2 - t1 = 5.83s - 1.75s = 4.08s.
    b. We find the speeds of the two rock from
        v1y = v01y+ gt1 = (20m/s) + (9.8m/s2)(1.75s) = 37.15m/s.
        v2y = v02y+ gt2 = (-20m/s) + (9.8m/s2)(5.83s) = 37.13m/s. Or the second rock returns the starting point with the same velocity as the first one does.
       So, they hit the water with the same speed.
  6. Both Jill and the cart accelerate toward to each other, the cart's acceleration is gsin3°.
    First we find the time elapse before Jill catches the cart.
        1/2aJillt2 + 1/2acartt2 = 1/2[(2m/s2) + 1/2acartt2]t2 = 50m, which gives t = 6.3s.
    We find the distance the cart rolls from
        x = 1/2acartt2 = 1/21/2acart(6.3s)2 = 10.2m.