Answer
With $f(t)$ = the difference in the positions of the 2 bicyclists and $[a, b]=[0,2.25]$
we use the Mean Value Theorem to prove that a time $t_{0}$ exists such that
$v_{1}(t_{0})=v_{2}(t_{0})$
(detailed proof in the "step-by-step" section.)
Work Step by Step
The Mean Value Theorem
If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ ,
then there exists a number $c$ in $(a, b)$ such that$ f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}.$
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Let $f(t)$ be the difference in the positions of the 2 bicyclists.
Let $[a, b]=[0,2.25]$
$f(t)=S_{1}(t)-S_{2}(t)$ .
We have:
$f(0)=0$
$f(2.25)=0$,
By the Mean Value Theorem, there must exist a time $t_{0}\in(0,2.25)$ such that
$f^{\prime}(t_{o})=S_{1}^{\prime}(t_{o})-S_{2}^{\prime}(t_{o})=0.$
At this time $t_0$ , since the derivative of position is velocity,
$v_{1}(t_{0})-v_{2}(t_{0})=0$
that is,
$v_{1}(t_{0})=v_{2}(t_{0})$.
