Question

A mass attached to a vertical spring has position function given by

Answer 1

a) -11.4 in/s

b) -29.3 in/s^2

Explanation:

We have the following function corresponding to the position at a given time:

`s(t)=5\sin (3t)`

If we differentiate this function with respect to time, we obtain the velocity function, like this:

`\begin{gathered} v(t)=s^(\prime)(t) \\ s^(\prime)(t)=(d)/(dt)(s(t)) \\ \frac{d}{\mathrm{d}t}(5\sin (3t))=15\cos (3t) \\ v(t)=15\cos (3t) \end{gathered}`

We calculate the velocity by replacing t = 5, just like this:

`\begin{gathered} v(t)=15\cos (3\cdot5)=15\cos (15) \\ v(t)=-11.4\text{ in/s} \end{gathered}`

If we differentiate the function again, we obtain the acceleration function, as follows:

`\begin{gathered} a(t)=v^(\prime)(t) \\ v^(\prime)(t)=(d)/(dt)(v(t)) \\ (d)/(dt)(15\cos (3t))=-45\sin \mleft(3t\mright) \\ a(t)=-45\sin (3t) \end{gathered}`

We calculate the acceleration by replacing t = 5, just like this:

`\begin{gathered} v(t)=-45\sin (3\cdot5)=-45\sin (15) \\ v(t)=-29.3in/s^2 \end{gathered}`