Question

A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the track is 28.2 m, what is the centripetal acceleration of the runner?
1.13 m/s2
0.284 m/s2
3.53 m/s2
2.27 m/s2

Answer 1

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

$a_c =w^2r = \frac{v^2}{r}$

in this case we know the speed of the runner

$v =8.00\ m/s$

The radius "r" will be the distance from the runner to the center of the track

$r = 28.2\ m$

$a_c = \frac{8^2}{28.2}\ m/s^2$

$a_c = 2.27\ m/s^2$

The answer is the last option