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.

## Expert Answer

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 a**ssuming** 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**.