Q. 74 - Calculus

Question

You slam on your brakes and come to a full stop exactly at a stop sign. Your distance from the sign after t seconds is $s\left(t\right) = 3t^3-12t^2-9t+54$ feet, as shown in the following graph:

(a) What does the Mean Value Theorem from Section 3.1 say about your distance and/or velocity from the stop sign during the time that you are applying your brakes?

(b) What does the Mean Value Theorem for Integrals say about your distance from the stop sign during the time that you are applying the brakes?

Step-by-Step Solution

Answer

(a) The rate of change of speed is -18 feet per second for some c in the first three seconds.

(b) The distance is s(c) = 24.75  for some c in the first three seconds.

Step 1. Given Information.

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

Step 2. part (a) The mean value theorem.

The mean value theorem says that there is some c in [0,3] for which 

$s\left(c\right) = \frac{s\left(3\right) - s\left(0\right)}{3-0},$

So,

$$\begin{gathered} s\left(3\right) = 3 {\left(3\right)}^3 - 12 {\left(3\right)}^2 -9\left(3\right) +54 \\ s\left(3\right) = 81-108-27+54 = 0. \\ s\left(0\right) = 0-0-0+54 = 54 \\ So, \\ s\left(3\right)-s\left(0\right) = 0-54 = -54 \\ Therefore, s\left(c\right) = \frac{-54}{3} = -18. \\ Therefore, the rate of change of speed is -18 feet per second \\ for some c in the first three seconds. \end{gathered}$$

Step 3. part (b) Finding the average distance.

$$\begin{gathered} Average dis\tan ce is: \\ =\frac{1}{3-0}{\int }_0^3(3t^3-12t^2-9t+54)dt \\ = \frac{1}{3}{\left(\frac{3t^4}{4}-\frac{12t^3}{3}-\frac{9t^2}{2}+54t\right)}_0^3 \\ = \frac{1}{3}\left(\frac{297}{4}\right) = 24.75 \\ Therefore, The dis\tan ce is s\left(c\right) = 24.75 for some c in the \\ first three seconds. \end{gathered}$$