Problem 6-11: Calculating the centripetal acceleration of a runner in a circular track

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 aca_{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 vv and has the magnitude

ac=v2ra_{c}=\frac{v^{2}}{r}

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

v=ΔxΔtv=\frac{\Delta x}{\Delta t}

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

v=ΔxΔtv=200 m23.2 sv=8.6207 m/s\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 mr=30\ \text{m} . We now have the details to solve for the centripetal acceleration.

ac=v2rac=(8.6207 m/s)230 mac=2.4772 m/s2ac=2.5 m/s2  (Answer)\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*}
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