ANSWER:
a) -11.4 in/s
b) -29.3 in/s^2
STEP-BY-STEP EXPLANATION:
We have the following function corresponding to the position at a given time:
[tex]s(t)=5\sin (3t)[/tex]
If we differentiate this function with respect to time, we obtain the velocity function, like this:
[tex]\begin{gathered} v(t)=s^{\prime}(t) \\ s^{\prime}(t)=\frac{d}{dt}(s(t)) \\ \frac{d}{\mathrm{d}t}(5\sin (3t))=15\cos (3t) \\ v(t)=15\cos (3t) \end{gathered}[/tex]
We calculate the velocity by replacing t = 5, just like this:
[tex]\begin{gathered} v(t)=15\cos (3\cdot5)=15\cos (15) \\ v(t)=-11.4\text{ in/s} \end{gathered}[/tex]
If we differentiate the function again, we obtain the acceleration function, as follows:
[tex]\begin{gathered} a(t)=v^{\prime}(t) \\ v^{\prime}(t)=\frac{d}{dt}(v(t)) \\ \frac{d}{dt}(15\cos (3t))=-45\sin \mleft(3t\mright) \\ a(t)=-45\sin (3t) \end{gathered}[/tex]
We calculate the acceleration by replacing t = 5, just like this:
[tex]\begin{gathered} v(t)=-45\sin (3\cdot5)=-45\sin (15) \\ v(t)=-29.3in/s^2 \end{gathered}[/tex]
