Q. 5 - Calculus

Question

You have just calculated your average velocity in two ways, once using the formula $\frac{s\left(b\right)-s\left(a\right)}{b-a}$ for the average rate of change of position and once using the definition $\frac{1}{b-a}{\int }_a^{b}v\left(t\right)dt$ of the average value of $v\left(t\right)$ on $\left[a,b\right]$, where $a=0$ and $b$ is the amount of time it took you to stop the car. Use the fact that these two quantities are equal to discuss the relationship between the area under the graph of your velocity $v\left(t\right)$ on $\left[a,b\right]$ and the total distance that you traveled while trying to stop the car.

Step-by-Step Solution

Answer

The area under the graph of the velocity is $3.4$ square feet and the total distance traveled is $50.6$feet.

The relationship is $S=A+54$.

Step 1. Given Information.

The function is,

$s\left(t\right)=3t^3-12t^2-9t+54$.

Step 2. Finding the relationship.

The area under the graph of the velocity on $\left[0,4.6\right]$ is,

$$\begin{gathered} v\left(t\right)=\frac{d\left(3t^3-12t^2-9t+54\right)}{dt} \\ =9t^2-24t-9 \end{gathered}$$

$$\begin{gathered} A\left(t\right)={\int }_0^{4.6}\left(9t^2-24t-9\right)dt \\ =9{\int }_0^{4.6}{t}^{2}dt-24{\int }_0^{4.6}tdt-9{\int }_0^{4.6}dt \\ =9{\left[\frac{t^3}{3}\right]}_0^{4.6}-24{\left[\frac{t^2}{2}\right]}_0^{4.6}-9{\left[t\right]}_0^{4.6} \\ =292-254-41.4 \\ =-3.4 \end{gathered}$$

The total distance traveled while trying to stop the car is,

$$\begin{gathered} s\left(t\right)=3t^3-12t^2-9t+54 \\ s(4.6)=3{\left(4.6\right)}^3-12{\left(4.6\right)}^2-9\left(4.6\right)+54 \\ s(4.6)=292-254-41.4-54 \\ s(4.6)=50.6 \end{gathered}$$

The relationship between the two is $S=A+54$.

The area under the graph of the velocity is  $3.4$ square feet and the total distance traveled is $50.6$ feet.