120k views
3 votes
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

User Migmaker
by
8.4k points

1 Answer

3 votes

Answer: Last option

2.27 m/s2

Step-by-step 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 = (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 = (8^2)/(28.2)\ m/s^2


a_c = 2.27\ m/s^2

The answer is the last option

User Robinnnnn
by
8.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.