Question

Suppose a merry-go-round is spinning once every three seconds if a point on the outside edge has a speed of 12.56 ft./s what is the radius of the merry-go-round

Answer 1

20. 
let the point be point P. point P travels linear speed = 12.56 ft/sec 
in three sec, or one revolution, it travels 

12.56 ft/sec times 3 sec= ......feet, which is the perimater of the imaginary circle. then use the perimter of circle formula, solve for radius

Answer 2

Answer:

6 feet

Step-by-step explanation:

Time = 3 seconds

Speed = 12.56 ft./s

$Distance = Speed \times Time$

$Distance = 12.56 \times 3$

$Distance = 37.68 feet$

So, circumference = 37.68 feet

Formula of Circumference  = $2 \pi r$

So, $2 \pi r = 37.68$

$2 \times 3.14 \times r = 37.68$

$r = \frac{37.68}{2  \times 3.14}$

$r =6$

Hence the radius of the merry-go-round is 6 feet