Question

The leader of a bicycle race is traveling with a constant velocity of +11.5 m/s and is 12.0 m ahead of the second-place cyclist. The second-place cyclist has a velocity of +9.10 m/s and an acceleration of +1.20 m/s2. How much time elapses before he catches the leader?

Answer 1

Question 35167

Let the coordinate system be placed at the position of the second cyclist. Then, coordinate of the leader of the race is $x_{1}(t) = 12 + v_{1}t = 12 + 11.5t$ . Since second cyclist is moving with acceleration $a = 1.2\frac{m}{s^2}$ , the coordinate of the second bicyclist is $x_{2}(t) = v_{2}^{0}t + \frac{at^{2}}{2} = 9.10t + \frac{1.2\cdot t^{2}}{2}$ . When second cyclist catches the leader, $x_{1}(t) = x_{2}(t)$ , which yields simple quadratic equation $9.10t + 0.6t^{2} = 12 + 11.5t$ . Solutions of the latter one are $t_1 = 6.9s; t_2 = -2.89s$ . Since time is positive, the second cyclist will reach the leader in 6.9 seconds.