Question

If the runner completes the 200 m dash in 23.2 s and runs at a constant speed around a circular arc with a radius of curvature of 30 m, what is the magnitude of their centripetal acceleration during the curved portion of the track?

a) 2.32 m/s²
b) 3.45 m/s²
c) 4.81 m/s²
d) 5.92 m/s²

Answer 1

The magnitude of the runner's centripetal acceleration during the curved portion of the track is approximately 2.48 m/s².

Step-by-step explanation:

The centripetal acceleration of an object moving in a circular path is given by the formula:

centripetal acceleration = (v²) / r

Where v is the linear velocity of the object and r is the radius of curvature.

In this case, the runner completes the 200 m dash in 23.2 seconds, so the linear velocity is:

v = d / t

= 200 / 23.2

= 8.62 m/s

Using the given radius of curvature of 30 m, we can calculate the centripetal acceleration:

centripetal acceleration = (8.62²) / 30

≈ 2.48 m/s²

Therefore, the magnitude of the runner's centripetal acceleration during the curved portion of the track is approximately 2.48 m/s².