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

6 feet

Explanation:

Time = 3 seconds

Speed = 12.56 ft./s

`Distance = Speed * Time`

`Distance = 12.56 * 3`

`Distance = 37.68 feet`

So, circumference = 37.68 feet

Formula of Circumference =
`2 \pi r`

So,
`2 \pi r = 37.68`

`2 * 3.14 * r = 37.68`

`r = (37.68)/(2  * 3.14)`

`r =6`

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