Here's how we find the distances at the specified times using the piecewise function [tex]\( D(t) \)[/tex].
1. Starting distance (at [tex]\( t = 0 \)[/tex] hours):
The function at [tex]\( t = 0 \)[/tex] is given by:
[tex]\[
D(0) = 300(0) + 125 = 125 \text{ miles}
\][/tex]
So, the starting distance is 125 miles.
2. Distance at [tex]\( t = 2 \)[/tex] hours:
For [tex]\( 0 \leq t < 2.5 \)[/tex], the distance function is [tex]\( D(t) = 300t + 125 \)[/tex]. Thus, at [tex]\( t = 2 \)[/tex]:
[tex]\[
D(2) = 300(2) + 125 = 600 + 125 = 725 \text{ miles}
\][/tex]
So, at 2 hours, the traveler is 725 miles from home.
3. Distance at [tex]\( t = 2.5 \)[/tex] hours:
For [tex]\( 2.5 \leq t \leq 3.5 \)[/tex], the distance function is [tex]\( D(t) = 875 \)[/tex]. Thus, at [tex]\( t = 2.5 \)[/tex]:
[tex]\[
D(2.5) = 875 \text{ miles}
\][/tex]
So, the traveler is not moving farther but has reached a constant distance of 875 miles at 2.5 hours.
4. Distance at [tex]\( t = 3 \)[/tex] hours:
For [tex]\( 2.5 \leq t \leq 3.5 \)[/tex], the distance function remains [tex]\( D(t) = 875 \)[/tex]. Thus, at [tex]\( t = 3 \)[/tex]:
[tex]\[
D(3) = 875 \text{ miles}
\][/tex]
So, the traveler's distance is constant at 875 miles at 3 hours.
5. Total distance from home after [tex]\( t = 6 \)[/tex] hours:
For [tex]\( 3.5 < t \leq 6 \)[/tex], the distance function is [tex]\( D(t) = 75t + 612.5 \)[/tex]. Thus, at [tex]\( t = 6 \)[/tex]:
[tex]\[
D(6) = 75(6) + 612.5 = 450 + 612.5 = 1062.5 \text{ miles}
\][/tex]
So, the total distance from home after 6 hours is 1062.5 miles.
Based on these calculations, the correct options are:
- At 2 hours, the traveler is 725 miles from home.
- At 3 hours, the distance is constant, at 875 miles.
- The total distance from home after 6 hours is 1062.5 miles.
