Chapter 14, Problem 17CT

An object is moving along a straight line according to some position function $s=s\left(t\right)$. The distance $s$(in feet) of the object, from its starting point after $t$ seconds is given in the table. Find the average rate of change of distance from $t=3$ to $t=6$ seconds.

$\begin{array}{cc}t& s\\ 0& 0\\ 1& 2.5\\ 2& 14\\ 3& 31\\ 4& 49\\ 5& 89\\ 6& 137\\ 7& 173\\ 8& 240\end{array}$

To determine

To calculate: Theaverage rate of change of distance from $t=3$ to $t=6$ seconds, where an object is moving along a straight line as per the position function $s=s\left(t\right)$. The distance $s$(in feet) of the object from its starting point after $t$ seconds is given in the following table,

$\begin{array}{cc}t& s\\ 0& 0\\ 1& 2.5\\ 2& 14\\ 3& 31\\ 4& 49\\ 5& 89\\ 6& 137\\ 7& 173\\ 8& 240\end{array}$

Answer to Problem 17CT

Solution:

The average rate of change of distance from $t=3$ to $t=6$ seconds is $35\frac{1}{3}$ feet per second or approximately $35.33$ feet per second.

Explanation of Solution

Given information:

An object is moving along a straight line as per the position function $s=s\left(t\right)$. The distance $s$(in feet) of the object from its starting point after $t$ seconds is given in the following table,

$\begin{array}{cc}t& s\\ 0& 0\\ 1& 2.5\\ 2& 14\\ 3& 31\\ 4& 49\\ 5& 89\\ 6& 137\\ 7& 173\\ 8& 240\end{array}$

Formula used:

Average rate of change from $c$ to $x,=\frac{f\left(x\right)-f\left(c\right)}{x-c}$, where $x\ne c$

Calculation:

By using the formula of average rate of change, Average rate of change of distance from $t=3$ to $t=6$ seconds, $=\frac{s\left(6\right)-s\left(3\right)}{6-3}$.

Since, from the given table the distance covered by an object after $6$ seconds is $s\left(6\right)=137$ feet and distance covered after $3$ seconds is $s\left(3\right)=31$ feet

By substituting $s\left(6\right)=137$ and $s\left(3\right)=31$ in $\frac{s\left(6\right)-s\left(3\right)}{6-3}$,

Average rate of change of distance from $t=3$ to $t=6$ seconds $=\frac{137-31}{6-3}$

$=\frac{106}{3}$

$=35\frac{1}{3}\approx 35.33$

Therefore, the average rate of change of distance from $t=3$ to $t=6$ seconds is $35\frac{1}{3}$ feet per second or approximately $35.33$ feet per second.