Final answer:
To calculate the child's acceleration on a merry-go-round, you must use the centripetal acceleration formula. Given a radius of 4.7 m and a time of one revolution of 6.6 seconds, the child's centripetal acceleration is approximately 1.353 m/s².
Step-by-step explanation:
To find the child's acceleration on a merry-go-round that makes one complete revolution in 6.6 seconds, we can use the concept of centripetal acceleration, which is given by the formula \( a = \frac{v^2}{r} \), where \( v \) is the linear velocity and \( r \) is the radius of the circular path.
First, we calculate the linear velocity (v) using the circumference of the circular path ( \( C = 2\pi r \) ) and the time for one revolution (T): \( v = \frac{C}{T} = \frac{2\pi r}{T} \).
Then, we plug the linear velocity back into the formula for centripetal acceleration:
\( a = \frac{(2\pi r)^2}{rT^2} = \frac{4\pi^2 r}{T^2} \).
In this case,
\( r = 4.7 \) meters, and
\( T = 6.6 \) seconds, so
\( a = \frac{4\pi^2 \cdot 4.7}{6.6^2} \approx \frac{4\cdot 3.1416^2 \cdot 4.7}{43.56} \approx \frac{58.9046}{43.56} \approx 1.353 \) m/s2.
Thus, the child's centripetal acceleration is approximately 1.353 m/s2.