Question

the distance in feet that a ball rolls down an incline after t seconds is modeled by the function s(t)=-3t² - 3. find the average velocity of the ball over the following time intervals: 2,2.1, 2,2.01, 2,2.001, and 2,2.0001. use this to draw a conclusion about the instantaneous velocity of the ball at t = 2 seconds. (round each answer to answer to 6 decimal places.) provide your answer below: t v_ave 2.1 2.01 2.001 2.0001 v_ave≈

Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{ave}$ over the interval $[a,b]$ for a position - function $s(t)$ is given by $v_{ave}=\frac{s(b)-s(a)}{b - a}$. Here, $a = 2$ and $s(t)=-3t^{2}-3$.

Step2: Calculate $s(2)$

Substitute $t = 2$ into $s(t)$: $s(2)=-3\times(2)^{2}-3=-3\times4 - 3=-12 - 3=-15$.

Step3: Calculate average velocity for $[2,2.1]$

First, find $s(2.1)=-3\times(2.1)^{2}-3=-3\times4.41-3=-13.23 - 3=-16.23$. Then, $v_{ave}=\frac{s(2.1)-s(2)}{2.1 - 2}=\frac{-16.23+15}{0.1}=\frac{-1.23}{0.1}=-12.300000$.

Step4: Calculate average velocity for $[2,2.01]$

Find $s(2.01)=-3\times(2.01)^{2}-3=-3\times4.0401-3=-12.1203 - 3=-15.1203$. Then, $v_{ave}=\frac{s(2.01)-s(2)}{2.01 - 2}=\frac{-15.1203 + 15}{0.01}=\frac{-0.1203}{0.01}=-12.030000$.

Step5: Calculate average velocity for $[2,2.001]$

Find $s(2.001)=-3\times(2.001)^{2}-3=-3\times4.004001-3=-12.012003 - 3=-15.012003$. Then, $v_{ave}=\frac{s(2.001)-s(2)}{2.001 - 2}=\frac{-15.012003+15}{0.001}=\frac{-0.012003}{0.001}=-12.003000$.

Step6: Calculate average velocity for $[2,2.0001]$

Find $s(2.0001)=-3\times(2.0001)^{2}-3=-3\times4.00040001-3=-12.00120003 - 3=-15.00120003$. Then, $v_{ave}=\frac{s(2.0001)-s(2)}{2.0001 - 2}=\frac{-15.00120003 + 15}{0.0001}=\frac{-0.00120003}{0.0001}=-12.000300$.
As the time - intervals get smaller and smaller around $t = 2$, the average velocities are approaching $- 12$. So, the instantaneous velocity at $t = 2$ is approximately $-12$.

Answer:

t	$v_{ave}$
2.01	-12.030000
2.001	-12.003000
2.0001	-12.000300

$v_{inst}\approx - 12$