# Two runners start a race at the same time and finish in a tie.

The main objective of this question is to prove that the two runners have the same speed during some interval of time in the race.

This question uses the concept of Calculus and Rolle’s theorem. In Rolle’s theorem, two conditions must be satisfied by a function that is defined in the interval [a,b]. The two conditions are that the given function must be differentiable and continuous in the open and closed interval respectively.

To prove that two runners have the same speed during the race at some interval of time, we are given:

$f(t) \space =\space g(t) \space – \space h(t)$

Where  $g(t)$  –   $h(t)$ is the difference in position bet ween two runners and  $g(t)$  and  $h(t)$ are continuous as well as differentiable which results  $f(t)$  continuous and differentiable. The $g(t)$ and $h(t)$ are the positions of two runners.

Taking the derivative of the given equation results in:

$\space f'(t) \space = \space g’=(t) \space – \space h'(t) \space$

Now assuming an interval $(t_0,t_1)$ for the  runners in the race.  The start time is $(t_0)$ while the $(t_1)$ is the finishing time. It is also given that the two runners start the race at the same time which results in finishing the race at the same time.

Then we have $(t_0) = h(t_0)$  and  $g(t_1) = h(t_1)$

Now we have:

$f(t_0) =0$ and  $f(t_1) =0$

This results allow us to use of Rolle’s theorem as $f(t_0) =f(t_1)$ and  $f(t_1) are differentiable as well as continuous. While$f^{‘}(c) = 0 $. So : $f'(c) \space = \space g'(c) \space – \space h'(c) \space = 0$ $g'(c) \space = \space h'(c)$ $c \space = \space t, \space t \space \in \space(t_0,t_1)$ $g'(t) \space = \space h'(t)$ Hence it is proved that the two runners in the race have the same speed during some interval of time. ## Numerical Answer By using the concept of Rolle’s theorem, it is proved that the two runners have the same speed at some interval of time during the race. ## Example Prove that two cars have the same speed during a race at some interval which results in finishing the race at the same time. By using the concept of Rolle’s theorem, we can prove that the two cars which finish the race at the same time have the same speed at some interval of time during the race. So we know that: $x(t) \space =\space y(t) \space – \space z(t)$ Where$y(t)$–$z(t)$is the difference in position bet ween two runners and$y(t)$and$z(t)$are continuous as well as differentiable which results$x(t)$continuous and differentiable. The derivative of the equation results in: $\space x'(t) \space = \space y'(t) \space – \space z'(t) \space$ Now assuming an interval$(t_0,t_1)$for the cars in the race. Then we have$(t_0)  =  z(t_0)$and$y(t_1)  =  z(t_1)x(t_0) =0$and$x(t_1) =0$This results allow us the use of Rolle’s theorem. While$x'(c) = 0 \$. So :

$x'(c) \space = \space y'(c) \space – \space z'(c) \space = 0$

$y'(c) \space = \space z'(c)$

$c \space = \space t, \space t \space \in \space(t_0,t_1)$

$y'(t) \space = \space z'(t)$

Hence, it is proved.