# Consider s(t)=5sin(t) describes a block bouncing on a spring. Complete the table with average velocities. Make the conjecture about the value of the instantaneous velocity at t=pi/2?

## [ $\frac{\pi}{2}$, $\pi$] [ $\frac{\pi}{2}$, $\frac{\pi}{2}$+0.1][ $\frac{\pi}{2}$, $\frac{\pi}{2}$+0.01]

Jun 13, 2017

1. ${v}_{\text{av}} = - 3.18$ $\text{m/s}$

2. ${v}_{\text{av}} = - 0.250$ $\text{m/s}$

3. ${v}_{\text{av}} = - 0.0250$ $\text{m/s}$

$v \left(\frac{\pi}{2}\right) = 0$ $\text{m/s}$

#### Explanation:

We're asked to calculate the average velocity for three different time intervals, and the instantaneous velocity at $t = \frac{\pi}{2}$ $\text{s}$.

1.

For the sfcolor(red)("first" average velocity, we'll plug in the two times $t = \frac{\pi}{2}$ $\text{s}$ and $t = \pi$ $\text{s}$ into the position equation to find the position at those times:

$s \left(\frac{\pi}{2}\right) = 5 \sin \left(\frac{\pi}{2}\right) = 5$ $\text{m}$

$s \left(\pi\right) = 5 \sin \left(\pi\right) = 0$ $\text{m}$

Next, we'll use the average velocity formula

${v}_{\text{av}} = \frac{\Delta x}{\Delta t}$

The quantity $\Delta x$ is the change in position of the object, which is

$0 \text{m" - 5"m" = -5"m}$.

(I'm assuming the distance units are meters.)

The quantity $\Delta t$ is the time interval, which is

$\pi$ $\text{s}$ $- \frac{\pi}{2}$ $\text{s} = \frac{\pi}{2}$ $\text{s}$

The average velocity is thus

v_"av" = (-5"m")/(pi/2"s") = -3.18 $\text{m/s}$

2.

Similarly, we'll plug in the times and find the position, then use this to find the average velocity:

$s \left(\frac{\pi}{2}\right) = 5 \sin \left(\frac{\pi}{2}\right) = 5$ $\text{m}$

$s \left(\frac{\pi}{2} + 0.1\right) = 5 \sin \left(\frac{\pi}{2} + 0.1\right) = 4.975$ $\text{m}$

$\Delta x = 0.025$ $\text{m}$

$\Delta t = 0.1$ $\text{s}$

${v}_{\text{av}} = \frac{\Delta x}{\Delta t} = - 0.250$ $\text{m/s}$

3.

$s \left(\frac{\pi}{2}\right) = 5 \sin \left(\frac{\pi}{2}\right) = 5$ $\text{m}$

$s \left(\frac{\pi}{2} + 0.01\right) = 5 \sin \left(\frac{\pi}{2} + 0.01\right) = 4.99975$ $\text{m}$

$\Delta x = - 0.000249998$ $\text{m}$

$\Delta t = 0.01$ $\text{s}$

${v}_{\text{av}} = - 0.0250$ $\text{m/s}$

4.

As we can see, if we keep plugging in smaller time intervals, the average velocity gets closer and closer to $0$. We can then conclude that the instantaneous velocity of the block at $t = \frac{\pi}{2}$ $\text{s}$ is color(red)(0 color(red)("m/s"