Question

A runner runs around a circular track. He completes one lap at a time of t = 314 s at a constant speed of v = 3.1 m/s. t = 314 sv = 3.1 m/sWhat is the radius, r in meters, of the track? What was the runners centripetal acceleration, ac in m/s2, during the run?

Answer 1

Since the runner completes 1 lap in 314 seconds, and its velocity is 3.1m/s, then the total distance covered in 1 lap is:

`\begin{gathered} d=vt \\ =(3.1(m)/(s))(314s) \\ =973.4m \end{gathered}`

That distance corresponds to the perimeter of the circumference. The perimeter of a circumference with radius r is 2πr. Then:

`\begin{gathered} 2\pi r=d \\ \\ \Rightarrow r=(d)/(2\pi) \\ =(973.4m)/(2(3.14...)) \\ =154.9...m \end{gathered}`

The centripetal acceleration of an object in a circular trajectory with radius r and speed v is:

`a_c=(v^2)/(r)`

Replace v=3.1m/s and r=154.9m to find the centripetal acceleration:

`a_c=((3.1(m)/(s))^2)/((154.9m))=0.062(m)/(s^2)`

Therefore, the radius of the track is approximately 155m and the centripetal acceleration of the runner is approximately 0.062 m/s^2.