[PHYSICS QUESTION] A girl rides a merry-go-round that rotates at a constant rate.
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[PHYSICS QUESTION] A girl rides a merry-go-round that rotates at a constant rate.

[From: ] [author: ] [Date: 11-09-18] [Hit: ]
we are told that 20m is 1/4 the circumference,remembering that C=2 pi r for a circle, we know that r=80/2 pi = 12.the linear velocity is then 20m/2.so the acceleration = 8^2/(12.74m) = 5.......
A girl rides a merry-go-round that rotates at a constant rate. She travels one fourth of a revolution, a distance of 20 m along the circumference of the circle, in 2.5 s. The magnitude of her acceleration is:

The answer is 5.0 m/s^2, but whatever I do I cannot arrive at that. How do I do this problem?

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acceleration in circular motion is v^2/r where v is the linear velocity and r is the radius of the circle

we are told that 20m is 1/4 the circumference, so the circumference is 80 m

remembering that C=2 pi r for a circle, we know that r=80/2 pi = 12.74m

the linear velocity is then 20m/2.5s = 8m/s

so the acceleration = 8^2/(12.74m) = 5.0m/s/s
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