To create a piecewise function that models the distance traveled by Julieta's family over time, we can break down the trip into two intervals based on their speed:
- First interval (0 to 1 hour): For the first hour, they travel at a speed of 55 mph.
- Second interval (1 to 3 hours): For the next two hours (from 1 hour to 3 hours), they travel at a speed of 70 mph.
Let's denote the distance \( D(t) \), where \( t \) is the time in hours.
Piecewise Function
The piecewise function can be defined as follows:
\[ D(t) = \begin{cases} 55t & \text{for } 0 \leq t < 1 \ 55 + 70(t - 1) & \text{for } 1 \leq t \leq 3 \end{cases} \]
Explanation
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For \( 0 \leq t < 1 \):
- The distance \( D(t) \) is calculated using the formula \( D(t) = 55t \), where \( t \) is the time in hours.
- Here, for every hour, they travel 55 miles.
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For \( 1 \leq t \leq 3 \):
- At \( t = 1 \), they have already traveled \( D(1) = 55 \times 1 = 55 \) miles.
- For \( t \) from 1 to 3, they continue traveling at 70 mph. The distance for this interval is calculated as: \[ D(t) = 55 + 70(t - 1) \]
- Here, \( t - 1 \) represents the time traveled since the second speed started (from hour 1 to hour \( t \)).
Final Function
Now, we can simplify the second part of the piecewise function:
\[ D(t) = \begin{cases} 55t & \text{for } 0 \leq t < 1 \ 55 + 70(t - 1) & \text{for } 1 \leq t \leq 3 \end{cases} \]
For the second case:
- Simplifying \( 55 + 70(t - 1) \): \[ D(t) = 55 + 70t - 70 = 70t - 15 \text{ for } 1 \leq t \leq 3 \]
Thus, the final piecewise function is:
\[ D(t) = \begin{cases} 55t & \text{for } 0 \leq t < 1 \ 70t - 15 & \text{for } 1 \leq t \leq 3 \end{cases} \]
This function accurately models the distance traveled by Julieta's family over time during their road trip.