The function [tex]\( D(t) \)[/tex] defines a traveler's distance from home, in miles, as a function of time, in hours.

[tex]\[
D(t) = \begin{cases}
300t + 125 & 0 \leq t \ \textless \ 2.5 \\
875 & 2.5 \leq t \leq 3.5 \\
75t + 612.5 & 3.5 \ \textless \ t \leq 6
\end{cases}
\][/tex]

Which times and distances are described by the function? Select three options.

A. The starting distance, at [tex]\( t = 0 \)[/tex].

B. At 2 hours, the traveler is _______ miles from home.

C. At 2.5 hours, the traveler is _______ miles from home.

D. At 3 hours, the traveler is _______ miles from home.

E. The total distance traveled is _______ miles.



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.