A merry-go-round completes one revolution in 40 s.What is the centripetal acceleration of a rider located at a distance 2 m from the center of the merry go-round? A) m/s² B) $\pi$ m/s² C)$\pi$ 2 m/s² D)$\pi$²/100 m/s² E)$\pi$² m/s²

Question

Quizplus · Accepted Answer

The formula for centripetal acceleration is: 
a = v^2/r 
where v is the speed and r is the radius of the circular path.

In this case, we are given that the merry-go-round completes one revolution (i.e. a 2* pi radians) in 40 seconds. Therefore, the speed of the rider is: 
v = (2* pi * r) / T 
where T is the time for one revolution,  which is 40s 
v = (2 * 3.14 * 2) / 40 = 0.314 m/s

Now, we can use the formula for centripetal acceleration: 
a = v^2 / r 
a = (0.314 m/s)^2 / 2m 
a = 0.049 m/s^2

Therefore, the correct answer is C, which gives the value of 0.049 m/s^2

A Merry-Go-Round Completes One Revolution in 40 S π\piπ M/s2 C) π\piπ

A Merry-Go-Round Completes One Revolution in 40 S $\pi$ M/s²
C) $\pi$