Answer :
To solve this problem, we need to evaluate the piecewise function [tex]\( D(t) \)[/tex] at the specified times to determine the traveler's distance from home. Here's how we can do it step-by-step:
### 1. The Starting Distance, at [tex]\( t = 0 \)[/tex]:
For [tex]\( t \)[/tex] in the interval [tex]\( 0 \leq t < 2.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is given by [tex]\( D(t) = 300t + 125 \)[/tex].
- When [tex]\( t = 0 \)[/tex]:
[tex]\[ D(0) = 300(0) + 125 = 125 \][/tex]
So, the starting distance is [tex]\( 125 \)[/tex] miles.
### 2. At 2 Hours, the Traveler Is:
For [tex]\( t \)[/tex] in the interval [tex]\( 0 \leq t < 2.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is given by [tex]\( D(t) = 300t + 125 \)[/tex].
- When [tex]\( t = 2 \)[/tex]:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \][/tex]
So, at 2 hours, the traveler is [tex]\( 725 \)[/tex] miles from home.
### 3. At 2.5 Hours, the Traveler:
For [tex]\( t \)[/tex] in the interval [tex]\( 2.5 \leq t \leq 3.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is a constant value:
[tex]\[ D(t) = 875 \][/tex]
- When [tex]\( t = 2.5 \)[/tex]:
[tex]\[ D(2.5) = 875 \][/tex]
So, at 2.5 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
### 4. At 3 Hours, the Distance:
For [tex]\( t \)[/tex] in the interval [tex]\( 2.5 \leq t \leq 3.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is again a constant value:
[tex]\[ D(t) = 875 \][/tex]
- When [tex]\( t = 3 \)[/tex]:
[tex]\[ D(3) = 875 \][/tex]
So, at 3 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
Combining these results, we have:
- The starting distance is [tex]\( 125 \)[/tex] miles.
- At 2 hours, the traveler is [tex]\( 725 \)[/tex] miles from home.
- At 2.5 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
- At 3 hours, the distance is [tex]\( 875 \)[/tex] miles.
Given these findings, the three noteworthy options are:
1. The starting distance, at [tex]\( t = 0 \)[/tex], is [tex]\( 125 \)[/tex] miles.
2. At 2 hours, the traveler is [tex]\( 725 \)[/tex] miles from home.
3. At 2.5 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
### 1. The Starting Distance, at [tex]\( t = 0 \)[/tex]:
For [tex]\( t \)[/tex] in the interval [tex]\( 0 \leq t < 2.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is given by [tex]\( D(t) = 300t + 125 \)[/tex].
- When [tex]\( t = 0 \)[/tex]:
[tex]\[ D(0) = 300(0) + 125 = 125 \][/tex]
So, the starting distance is [tex]\( 125 \)[/tex] miles.
### 2. At 2 Hours, the Traveler Is:
For [tex]\( t \)[/tex] in the interval [tex]\( 0 \leq t < 2.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is given by [tex]\( D(t) = 300t + 125 \)[/tex].
- When [tex]\( t = 2 \)[/tex]:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \][/tex]
So, at 2 hours, the traveler is [tex]\( 725 \)[/tex] miles from home.
### 3. At 2.5 Hours, the Traveler:
For [tex]\( t \)[/tex] in the interval [tex]\( 2.5 \leq t \leq 3.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is a constant value:
[tex]\[ D(t) = 875 \][/tex]
- When [tex]\( t = 2.5 \)[/tex]:
[tex]\[ D(2.5) = 875 \][/tex]
So, at 2.5 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
### 4. At 3 Hours, the Distance:
For [tex]\( t \)[/tex] in the interval [tex]\( 2.5 \leq t \leq 3.5 \)[/tex], the function [tex]\( D(t) \)[/tex] is again a constant value:
[tex]\[ D(t) = 875 \][/tex]
- When [tex]\( t = 3 \)[/tex]:
[tex]\[ D(3) = 875 \][/tex]
So, at 3 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
Combining these results, we have:
- The starting distance is [tex]\( 125 \)[/tex] miles.
- At 2 hours, the traveler is [tex]\( 725 \)[/tex] miles from home.
- At 2.5 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
- At 3 hours, the distance is [tex]\( 875 \)[/tex] miles.
Given these findings, the three noteworthy options are:
1. The starting distance, at [tex]\( t = 0 \)[/tex], is [tex]\( 125 \)[/tex] miles.
2. At 2 hours, the traveler is [tex]\( 725 \)[/tex] miles from home.
3. At 2.5 hours, the traveler is [tex]\( 875 \)[/tex] miles from home.
