Answers to Chapter 2

1. (a) We find the elapsed time before the speed change from

                   speed = d₁/t₁;

                   65mi/h = (130mi)/t₁, which gives t₁ = 2.0h.

              Thus the time at the lower speed is

                   t₂ = T - t₁ = 3.33h- 2.0h = 1.33h.

              We find the distance traveled at the lower speed from

                   speed = d₂/t₂;

                   55mi/h = d₂/(1.33h), which gives d₂ = 73mi.

               The total distance traveled is

                   D = d₁ + d₂ = 130mi + 73mi = 203mi.

           (b) We find the average speed from

                   average speed = d/t = (203mi)/(3.33h) =  61mi/h.

                 Note  that the average speed is not 1/2(65mi + 55mi). The two speeds were not maintained for equal times.

2. We find the time for the outgoing 200km from

               t₁ = d₁/v₁ = (200km)/(90km/h) = 2.22h.

        We find the time for the return 200km from

               t₂ = d₂/v₂ = (200km)/(50km/h) = 4.00h.

         We find the average speed from

              average speed = (d₁ + d₂)/(t₁ + t_lunch + t₂)

                                     = (200km + 200km)/ (2.22h + 1.00h + 4.00h) = 55km/h.

         Because the trip finishes at the starting point, there is no displacement; thus the average velocity is

               v_av = Dx/Dt = 0.

3. We find the acceleration (assumed to be constant) from

              v² = v₀² + 2a(x₂ - x₁);

              0 =[(90km/h)/(3.6ks/h)]² + 2a(50m), which gives  a = -6.3m/s².

         The number of g's is

              N = |a|/g = (6.3m/s²)/(9.80m/s²) = 0.64.

4. We convert the units for the speed: (45km/h)/(3.6ks/h) = 12.5m/s.

        (a) We find the distance the car travels before stopping from

                 v² = v₀² + 2a(x₁ - x₀);

                  0  = (12.5m/s)² + 2(-0.50m/s²))(x₁ - x₀), which gives&nbbsp; x₁ - x₀ = 1.6 x 10²m.

         (b) We find the time it takes to stop the car from

                   v = v₀ + at;

                   0 = 12.5m/s + (-0.50m/s²)t, which gives  t = 25s.

          (c) With the origin at the beginning of the coast, we find the position at a time t from

                   x = v₀t + 1/2at². thus we find

                    x₁ = (12.5m/s)(1.0s) + 1/2(-0.50m/s²)(1.0s)² = 12m;

                    x₄ = (12.5m/s)(4.0s) + 1/2(-0.50m/s²)(4.0s)² = 46m;

                    x₅ = (12.5m/s)(5.0s) + 1/2(-0.50m/s²)(5.0s)² = 56m.

               During the first second the car travels 12m - 0 = 12m.

               During the fifth second the car travels 56m - 46m = 10m.

5. We use a coordinate system with the origin at the ground and up positive.

        (a) At the top of the motion the velocity is zero, so we find the height h from 

                v² = v₀² +2ah;      

                0 = (25m/s)² + 2(-9.80m/s²)h, which give  h = 32m.

        (b) When the ball returns to the ground, its displacement is zero, so we have 

                y = y₀ + v₀t + 1/2at²

               0 = 0 + (25m/s)t + 1/2(-9.80m/s²))t²,

              which give t = 0 (when the ball starts up), and   t = 5.1s.

6. We use a coordinate system with origin at the ground and up positive.

        (a) We find the velocity from

                v² = v₀² + 2a(y - y₀);

                v² = (20.0m/s)² + 2(-9.80m/s²)(12.0m -0), which give v = +12.8m/s.

         The stone reaches this height on the way up (the positive sign) and on the way down (the negative sign).

        (b) we find the time to reach the height from

                v = v₀ + at;

               +12.8m/s = 20.0m/s + (-9.80m/s²)t, which gives t = 0.735s, 3.35s.

         (c) there are two answers because the stone reaches this height on the way up (t = 0.735s) and on the way down (t = 3.35s).