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Mamont248 [21]
4 years ago
12

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
Physics
1 answer:
Genrish500 [490]4 years ago
3 0

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

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