To solve this problem, we need to understand how the distances for Annabeth and Charlie change over time.
1. Annabeth's Distance Equation Analysis:
Annabeth's distance from New York is given by the function:
[tex]\[ D(t) = 60|t-3| \][/tex]
This equation describes a V-shaped graph with the vertex (the lowest point) at [tex]\( t = 3 \)[/tex]. Prior to [tex]\( t = 3 \)[/tex], the distance decreases as time approaches 3, and after [tex]\( t = 3 \)[/tex], the distance increases as time moves away from 3. Therefore, Annabeth's distance stops decreasing and starts increasing at [tex]\( t = 3 \)[/tex] hours. Thus, Annabeth travels for [tex]\( 3 \)[/tex] hours before her distance starts increasing.
2. Charlie's Distance Table Analysis:
Charlie's distance from New York is provided in the table:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline t & 0 & 2 & 4 & 6 & 8 & 10 \\
\hline F(t) & 270 & 180 & 90 & 0 & 90 & 180 \\
\hline
\end{array}
\][/tex]
From this table, we observe the following distances at various times:
- At [tex]\( t = 0 \)[/tex] hours, Charlie is 270 miles away.
- At [tex]\( t = 2 \)[/tex] hours, Charlie is 180 miles away.
- At [tex]\( t = 4 \)[/tex] hours, Charlie is 90 miles away.
- At [tex]\( t = 6 \)[/tex] hours, Charlie is 0 miles away.
- At [tex]\( t = 8 \)[/tex] hours, Charlie is 90 miles away again.
- At [tex]\( t = 10 \)[/tex] hours, Charlie is 180 miles away again.
From the table, the distance [tex]\( F(t) \)[/tex] is decreasing until [tex]\( t = 6 \)[/tex] hours and then starts increasing. So, Charlie's distance stops decreasing and starts increasing at [tex]\( t = 6 \)[/tex] hours. Thus, Charlie travels for [tex]\( 6 \)[/tex] hours before his distance starts increasing.
Conclusion:
Annabeth travels for [tex]\( 3 \)[/tex] hours, while Charlie travels for [tex]\( 6 \)[/tex] hours before their distances from New York stop decreasing and start increasing. Therefore, Charlie has travelled for a greater amount of time before his distance starts increasing.
The correct answer is:
A. Charlie
