Answer:
We can use the following kinematic equation to solve the problem:
Δx = v1(t) + (1/2)a2t²
where Δx is the initial separation between the two cyclists, v1 is the velocity of the leader, a2 is the acceleration of the second-place cyclist, and t is the time taken for the second-place cyclist to catch up with the leader.
Substituting the given values, we get:
9.60 m = (11.1 m/s)t + (1/2)(1.20 m/s²)t²
Simplifying and rearranging the equation, we get a quadratic equation in t:
0.6t² + 11.1t - 9.60 = 0
We can solve for t using the quadratic formula:
t = [-11.1 ± √(11.1² - 4(0.6)(-9.60))] / (2(0.6))t = [-11.1 ± 12.7] / 1.2
t = 0.91 s or t = -18.43 s
Since time cannot be negative, we can discard the negative solution. Therefore, the time taken for the second-place cyclist to catch up with the leader is:
t = 0.91 s
Hence, it will take 0.91 s for the second-place cyclist to catch up with the leader.
Step-by-step explanation: