Question

A child is playing with a yo-yo. When the child does a certain trick, the yo-yo's position in feet relative to the ground is S(t)=−1.1sin(2πt)+3.25
, where t
is the time in seconds.

After how many seconds, to the nearest thousandth of a second, is the yo-yo first at 2.65
feet?



0.092 seconds
0.577 seconds
1.154 seconds
0.046 seconds

Answer 1

To find the time \( t \) when the yo-yo is first at 2.65 feet, we need to solve the equation given by the position function:

\[
S(t) = -1.1 \sin(2\pi t) + 3.25
\]

We want to set \( S(t) \) equal to 2.65:

\[
-1.1 \sin(2\pi t) + 3.25 = 2.65
\]

First, let's isolate the sine term:

\[
-1.1 \sin(2\pi t) = 2.65 - 3.25
\]
\[
-1.1 \sin(2\pi t) = -0.60
\]

Now divide both sides by -1.1:

\[
\sin(2\pi t) = \frac{-0.60}{-1.1} = \frac{0.60}{1.1} \approx 0.545454545
\]

Next, we will take the arcsine (inverse sine) to find \( 2\pi t \):

\[
2\pi t = \arcsin(0.545454545)
\]

Using a calculator to find \( \arcsin(0.545454545) \):

\[
2\pi t \approx 0.577
\]

Now, we find \( t \):

\[
t \approx \frac{0.577}{2\pi} \approx \frac{0.577}{6.28318} \approx 0.092
\]

So, the first time the yo-yo is at 2.65 feet is approximately:

\[
t \approx 0.092 \, \text{seconds}
\]

Thus, the answer is:

\[
\boxed{0.092} \, \text{seconds}
\]