Question

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?

[ `pi/2`, `pi`] [ `pi/2`, `pi/2`+0.1][ `pi/2`, `pi/2`+0.01]

Answer 1

1. `v_"av" = -3.18` `"m/s"`

2. `v_"av" = -0.250` `"m/s"`

3. `v_"av" = -0.0250` `"m/s"`

`v(pi/2) = 0` `"m/s"`

Explanation:

We're asked to calculate the average velocity for three different time intervals, and the instantaneous velocity at `t = pi/2` `"s"`.

1.

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

`s(pi/2) = 5sin(pi/2) = 5` `"m"`

`s(pi) = 5sin(pi) = 0` `"m"`

Next, we'll use the average velocity formula

`v_"av" = (Deltax)/(Deltat)`

The quantity `Deltax` is the change in position of the object, which is

`0"m" - 5"m" = -5"m"`.

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

The quantity `Deltat` is the time interval, which is

`pi` `"s"` `- pi/2` `"s" = pi/2` `"s"`

The average velocity is thus

`v_"av" = (-5"m")/(pi/2"s") = -3.18` `"m/s"`

2.

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

`s(pi/2) = 5sin(pi/2) = 5` `"m"`

`s(pi/2 + 0.1) = 5sin(pi/2 + 0.1) = 4.975` `"m"`

`Deltax = 0.025` `"m"`

`Deltat = 0.1` `"s"`

`v_"av" = (Deltax)/(Deltat) = -0.250` `"m/s"`

3.

`s(pi/2) = 5sin(pi/2) = 5` `"m"`

`s(pi/2 + 0.01) = 5sin(pi/2 + 0.01) = 4.99975` `"m"`

`Deltax = -0.000249998` `"m"`

`Deltat = 0.01` `"s"`

`v_"av" = -0.0250` `"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 = pi/2` `"s"` is `color(red)(0` `color(red)("m/s"`