To help us solve this problem let's plot the points given in the table:
From the graph we notice that this the position can be modeled by a sine function, we also notice that the period of this function is 8. We know that a sine function can be modeled by:
[tex]A\sin(B(x+C))+D[/tex]
where A is the amplitude, C is the horizontal shift, D is the vertical shift and
[tex]\frac{2\pi}{B}[/tex]
is the period.
From the graph we have we notice that we don't have any horizontal or vertical shift, then C=0 and D=0. We also notice that the amplitude is 15, then A=15. Finally, as we said, the period is 8, then:
[tex]\begin{gathered} 8=\frac{2\pi}{B} \\ B=\frac{2\pi}{8} \\ B=\frac{\pi}{4} \end{gathered}[/tex]
Plugging these values in the sine function we have:
[tex]x(t)=15\sin(\frac{\pi}{4}t)[/tex]
If we graph this function along the points on the table we get the following graph:
We notice that we don't get an exact fit but we get a close one.
Now, that we have a function that describes the position we can find the velocity by taking the derivative:
[tex]\begin{gathered} x^{\prime}(t)=\frac{d}{dt}\lbrack15\sin(\frac{\pi}{4}t)\rbrack \\ =\frac{15\pi}{4}\cos(\frac{\pi}{4}t) \end{gathered}[/tex]
Therefore, the velocity is:
[tex]x^{\prime}(t)=\frac{15\pi}{4}\cos(\frac{\pi}{4}t)[/tex]
Once we have the expression for the velocity we can find values for the times we need, they are shown in the table below:
From the table we have that:
[tex]x^{\prime}(0.5)=10.884199\text{ cm/s}[/tex]
And that:
• The earliest time when the velocity is zero is 2 s.
,
• The second time when the velocity is zero is 6 s.
,
• The minimum velocity happens at 4 s.
,
• The minimum velocity is -11.780972 cm/s
