To solve part (b) of the problem, we need to find the average rate of change for the distance driven from 5 seconds to 9 seconds.
Step 1: Identify the given data points for time and distance.
- At [tex]\( t = 5 \)[/tex] seconds, the distance [tex]\( D(t) = 151.5 \)[/tex] meters.
- At [tex]\( t = 9 \)[/tex] seconds, the distance [tex]\( D(t) = 255.9 \)[/tex] meters.
Step 2: Use the formula for average rate of change. The average rate of change of a function [tex]\( D(t) \)[/tex] over an interval from [tex]\( t = a \)[/tex] to [tex]\( t = b \)[/tex] is given by:
[tex]\[
\text{Average Rate of Change} = \frac{D(b) - D(a)}{b - a}
\][/tex]
In this problem, [tex]\( a = 5 \)[/tex] and [tex]\( b = 9 \)[/tex].
Step 3: Substitute the given values into the formula.
[tex]\[
\text{Average Rate of Change} = \frac{D(9) - D(5)}{9 - 5} = \frac{255.9 \text{ meters} - 151.5 \text{ meters}}{9 \text{ seconds} - 5 \text{ seconds}}
\][/tex]
Step 4: Calculate the differences.
[tex]\[
\text{Distance Difference} = 255.9 \text{ meters} - 151.5 \text{ meters} = 104.4 \text{ meters}
\][/tex]
[tex]\[
\text{Time Difference} = 9 \text{ seconds} - 5 \text{ seconds} = 4 \text{ seconds}
\][/tex]
Step 5: Compute the average rate of change.
[tex]\[
\text{Average Rate of Change} = \frac{104.4 \text{ meters}}{4 \text{ seconds}} = 26.1 \text{ meters per second}
\][/tex]
Thus, the average rate of change for the distance driven from 5 seconds to 9 seconds is [tex]\(\boxed{26.1}\)[/tex] meters per second.
