Question

Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed.

Short answer

It is proved that at some time during the race they have the same speed.

Step-by-step solution

The mean value theorem

Let \(f\) be a function that is continuous on the closed interval \(\left( {a,b} \right)\) and differentiable on the open interval \(\left( {a,b} \right)\). Then there is a number \(c \in \left( {a,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( b \right)}}{{b - a}}\).

The velocity of the object

The derivative of the object at any time is the velocity of the object.

Let \(g\left( t \right)\) and \(h\left( t \right)\) be the position functions of the two runners and let \(f\left( t \right) = g\left( t \right) - h\left( t \right)\). Suppose, the finishing time of the race is \(b\).

Now, the starting position of the runners is same. So, \(g\left( 0 \right) = h\left( 0 \right) \Rightarrow f\left( 0 \right) = 0\).

The finishing position of the runners is also same. So, \(g\left( b \right) = h\left( b \right) \Rightarrow f\left( b \right) = 0\).

Now, using the Mean Value Theorem on the function \(f\left( t \right)\) in the interval \(\left( {0,b} \right)\), there exists a number \(c\) such that

\(\begin{aligned}{c}f'\left( c \right) &= \frac{{f\left( b \right) - f\left( 0 \right)}}{{b - 0}}\\ &= \frac{{0 - 0}}{b}\\ &= 0\end{aligned}\)

That implies at any time \(c\), \(f'\left( c \right) = 0\). Since, \(f'\left( c \right) = f'\left( c \right) - g'\left( c \right)\). So, \(f'\left( c \right) = g'\left( c \right)\).

Hence, it is proved that at some point of time, the velocities of the two runners are same.