Stadium Paradox

Paradox
Consider three rows of bodies:


 * $\begin{array} {ccccc}

\text {(A)} & 0 & 0 & 0 & 0 \\ \text {(B)} & 0 & 0 & 0 & 0 \\ \text {(C)} & 0 & 0 & 0 & 0 \\ \end{array}$

Let row $\text {(A)}$ be at rest, while row $\text {(B)}$ and row $\text {(C)}$ are travelling at the same speed in opposite directions.


 * $\begin{array} {ccccccc}

\text {(A)} &  & 0 & 0 & 0 & 0 & \\ \text {(B)} & 0 & 0 & 0 & 0 & & \\ \text {(C)} &  &   & 0 & 0 & 0 & 0 \\ \end{array}$

By the time they are all in the same part of the course, $\text {(B)}$ will have passed twice as many of the bodies in $\text {(C)}$ as $\text {(A)}$ has.

Therefore the time it takes to pass $\text {(A)}$ is twice as long as it takes to pass $\text {(C)}$.

But the time which $\text {(B)}$ and $\text {(C)}$ take to reach the position of $\text {(A)}$ is the same.

Therefore double the time is equal to half the time.