Question

Review. You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 49.0 mi/h with constant acceleration 8.00 mi/h/s and thereafter moves with a constant speed of 49.0 mi/h. At the same time, the cyclist speeds up from rest to 23.0 mi/h with constant acceleration 12.00 mi/h/s and thereafter moves with a constant speed of 23.0 mi/h. [HINT: this is a great problem to use some graphs to help you visualize what is going on. There are more than 2 important clock readings in this problem.] (a) For what time interval (in s) after the light turned green is the bicycle ahead of your car?

Answer 1

Initial velocity of car (u_c) = 0 a = 8.00 mi/h/s = 3.58 m/s^2 final velocity (v_c) = 49.0 mi/h = 21.92 m/s tie time take for the car to reach velocity 49.0 mi/h t_c = v_c - u_c/a_c = 49 mi/h - 0/8 mi/h/s = 6.13 sec let the time = t the bicycle is ahead of car so total distance covered by car S_t = 1/2 a_c t^2_c + v_c (t - 6.23) S_c similarly initial velocity of bicycle (u_b) = 0 a_b = 12 mi/h/s = 5.36 m/s^2 v_b = 23 mi/h = 10.28 m/s t_b = v_b - u_b/a_b = 10.28 m/s - 0/5.36 = 1.92 sec total distance covered by bicycle S_b = 0.5 times a_b t^2_b + v_b (t - t_b) bicycle is ahead of car when 1/2 a_c t^2 = 0.5 times a_b t^2_b + v_b (t - t_b) 0.5 times 3.58 m/s^2 times t^2 = 0.5 times 5.36 times 1.92^2 + 10.28 (t - 1.92) 1.79 t^2 = 9.88 +