A speeder traveling at a constant speed of 121 km/h races past a billboard. A patrol car pursues from rest with constant acceleration of (8.3 km/h)/s until it reaches its maximum speed of 201 km/h, which it maintains until it catches up with the speeder. (a) How long does it take the patrol car to catch the speeder if it starts moving just as the speeder passes? (in s) (b) How far does each car travel? (in km) Here are the equations I have been using: Vx=V_ox + (a_x)(t) X=vt+(1/2)at^2 and the special case... X-x_o=vt+(1/2)at^2 I first converted 121 km/h to 33.61 m/s, 201 km/h to 55.83 m/s and acceleration of 8.3 km/h/s to 2.3056 m/s^2 Using Vx=v_ox+(a_x)(t) I found the time it took to reach a velocity of 55.83 m/s (since v_ox is 0 m/s at acceleration is 2.3056 m/s^2) to be 24.21 s. In 24.21 s, the patrol car (car 2) has traveled 675.96 m. (using the distance formula)..(1/2)(2.3056)(24.21)^2 In that same time, car 1(the speeder) has traveled 813.6981 m. (33.61 m/s * 24.21 s) Next, I used the special case X-x_o=vt+(1/2)at^2 Car 1 x-813.6981 m = 33.61 m/s *t Car 2 x-675.96 m = (1/2)(2.3056 m/s^2)*t^2 X=33.61t + 813.6981 X=1.1528t^2+675.96 Since X is common, I can set these two equations equal to one another. 1.1528t^2-33.61t-137.738=0 t can now be found using the quadratic equation a=1.1528, b=-33.61, c=-137.738 I get two answers: 32.797, -3.642 (the second is of course nonsensical) When I plug back into special case eqs (X=33.61t+813.6981 and 1.1528t^2+675.96) I get 1916.00 m and 1915.96 m which = (roughly) 1.9 km. 33s until the officer reaches the speeder and it takes him 1.9 km to do this. I believe I have the correct work, but might have a rounding error? The answer in the back of the book (which has similar numbers) is close to what I have as an answer...But when I tried my method using the book's numbers, I didn't get the same answer (although I was close). Help please? THANK YOU!