Question

Cheryl is riding on the edge of a merry-go-round, 2m from the center, which is rotating with an increasing angular speed. Cheryl’s tangential acceleration is 3.0m/s2. At the instant that Cheryl’s linear speed is 4.0m/s, what is Cheryl’s total acceleration?

Answer 1

Answer:

$a=8.5m/s^2$

Explanation:

The total acceleration will be the vectorial sum of the tangential and centripetal accelerations, since they would be the only ones acting on Cheryl. They are perpendicular to each other, so the magnitude of the sum is given by:$a=\sqrt{a_{t}^2+a_{cp}^2}$

Since the centripetal acceleration is given by the formula $a_{cp}=\frac{v^2}{r}$, we have:

$a=\sqrt{a_{t}^2+(\frac{v^2}{r})^2}$

Which for our values is:

$a=\sqrt{(3m/s^2)^2+(\frac{(4m/s)^2}{(2m)})^2}=8.5m/s^2$