A merry go round at the park is spinning. You stand on the outer edge and are traveling at constant speed of 5 m/s. You walk halfway toward the center. What happened to your acceleration?

a' = 2a

Explanation:

The acceleration during this circular motion is the centripetal acceleration. Initially, the acceleration will be given as:

a = v²/r   equation (1)

Where,

a = centripetal acceleration at outer edge

v = speed of merry go round = 5 m/s

r = distance from center of round to person (outer edge)

Now, when the person moves half way towards center. Then, the acceleration becomes:

a' = v²/r'

where,

r' = r/2

Therefore,

a' = v²/(r/2)

a' = 2v²/r

using equation (1):

a' = 2a

i think the answer is a and b

i wish you luck

the ima (higher than 1m) will be greater than the ima with the height of 1m.