Question

Two cyclists are in a race. One cyclist knows that he is slower, so he cheats: he removes the faster cyclist’s bike chain. The cheater starts from rest immediately with acceleration 2.1 m/s2. The faster cyclist has to take 4 seconds to replace her bike chain. She then follows (also from rest) with acceleration 2.9 m/s2. Assume that both cyclists accelerate smoothly and that they do not reach their maximum speeds during this race.

What is the maximum length that the race can be (in meters) in order for the slower cyclist to win?

Answer 1

\[x_{slow}=x_{0}+v _{0}t+\frac{ 1 }{ 2 }at ^{2}\] and \[x _{fast}=x _{0}+v _{0}(t-4)+\frac{ 1 }{ 2 }a _{2}(t-4)^{2}\] we have \[x _{0}=0\] and \[v _{0}=0\] so \[x _{slow}=\frac{ 1 }{ 2 }a_{1}t ^{2}\] and \[x _{fast}=\frac{ 1 }{ 2 }a_{2}(t-4)^{2}\] the maximum lenght is when \[x _{slow}=x _{fast}\] so \[\frac{ 1 }{ 2 }a _{1}t ^{2}=\frac{ 1 }{ 2 }a _{2}(t-4)^{2}\] we got \[t=8,33\] and we plug it in the first or the second equation (slow or fast) \[x=\frac{ 1 }{ 2 }\times 2,1\times (8,33^{2})\] and the result is \[x=72,85\]