A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If the runner completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of their centripetal acceleration as they run the curved portion of the track?

Solution:

Centripetal acceleration a_{c} is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude

a_{c}=\frac{v^{2}}{r}

We can solve for the constant speed of the runner using the formula

v=\frac{\Delta x}{\Delta t}

We are given the distance \Delta x = 200 \ \text{m} , and the total time \Delta t = 23.2\ \text{s} . Therefore, the velocity is

\begin{align*}
v & =\frac{\Delta x}{\Delta t} \\ \\ 
v & = \frac{200\ \text{m}}{23.2\ \text{s}} \\ \\
v & = 8.6207\ \text{m/s}
\end{align*}

From the given problem, we are given the following values: r=30\ \text{m} . We now have the details to solve for the centripetal acceleration.

\begin{align*}
a_{c} & = \frac{v^{2}}{r} \\ \\
a_{c} & = \frac{\left( 8.6207\ \text{m/s} \right)^2}{30\ \text{m}} \\ \\
a_{c} & = 2.4772\ \text{m/s}^{2} \\ \\
a_{c} & = 2.5\  \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}