Problem 73 - Contemporary Precalculus

Question

Suppose a car travels at a constant rate of 55 mph for 2 hours and travels at 45 mph thereafter. Show that distance traveled is a function of time, and find the rule of the function.

Step-by-Step Solution

Answer

Answer: The rule of the function is given by $d(t) = \begin{cases} 55t & \text{for } 0 \leq t \leq 2 \\ 45t + 20 & \text{for } t > 2 \end{cases}$

Step 1: Analyze the given information

We are given that the car travels at a constant rate of 55 mph for the first 2 hours, and at a constant rate of 45 mph for the rest of the journey. We can find the distance traveled during each part of the journey using the formula Distance = Speed × Time. We'll need to split the problem into two parts: when the time is less than or equal to 2 hours and when the time is greater than 2 hours.

Step 2: Calculate distance traveled during the 55 mph leg

For the part of the journey where the car is traveling at 55 mph, the distance traveled can be calculated as: \(distance1 = 55t\), for \(0 \leq t \leq 2\), where \(t\) is the time in hours.

Step 3: Calculate distance traveled during the 45 mph leg

For the part of the journey where the car is traveling at 45 mph, the distance traveled can be calculated as: \(distance2 = 45(t-2)\), where \(t\) is the time in hours and we subtract 2 from the time to account for the time spent during the first leg of the journey. This expression for distance2 is valid when \(t > 2\).

Step 4: Combine the distances to form the function

Now we need to combine the expressions for distance1 and distance2 into a single function. We can do this by using a piecewise function since the speed changes at a specific point in time (i.e., 2 hours). The function we are looking for is: $d(t) = \begin{cases} 55t & \text{for } 0 \leq t \leq 2 \\ 110 + 45(t-2) & \text{for } t > 2 \end{cases}$

Step 5: Simplify the function

We can simplify the function as follows: $d(t) = \begin{cases} 55t & \text{for } 0 \leq t \leq 2 \\ 110 + 45t - 90 & \text{for } t > 2 \end{cases} $ which simplifies to $d(t) = \begin{cases} 55t & \text{for } 0 \leq t \leq 2 \\ 45t + 20 & \text{for } t > 2 \end{cases}$ This is the rule of the function describing the distance traveled as a function of time.

Key Concepts

Distance as a Function of Time

Understanding how distance relates to time is a crucial concept in mathematics and physics. Here, we explore this by examining how a car's distance traveled can be expressed as a function of time. When a car travels at a constant speed, the distance covered can be simply calculated by multiplying speed by time. This forms a straightforward linear relationship.
To illustrate, consider a car that travels at 55 mph. For the first 2 hours, the distance would be represented as \[ ext{Distance} = 55 imes t \], where \( t \) is time in hours. This means that for every hour, the car travels an additional 55 miles. For later hours, we have another rate of travel which requires thoughtful consideration when expressing distance over a given time interval.

Speed and Distance Calculations

Accurate calculations of speed and distance are essential in modeling real-world movement using math. For any segment where a constant speed is maintained, calculating distance is straightforward. Speed is simply the rate of change of distance over time.

• For the first leg at 55 mph, covering 2 hours, the formula becomes \[ ext{Distance} = 55 imes t \].
• For the second leg at 45 mph, active after 2 hours, the formula adjusts because the time reference needs adapting. The proper expression is \[ ext{Distance} = 45 imes (t-2) \], taking into account the initial 2 hours at a different speed.

By structuring the use of these formulas correctly, you can distinguish the role that different speeds play at different stages of the journey.

Piecewise Function Notation

Piecewise function notation is a powerful method for representing a scenario where different conditions apply, such as segments of a journey. Here, the function is split depending on the value of the independent variable, often time.
For the situation where a car transitions from 55 mph to 45 mph, we use piecewise notation to clearly define each segment of travel:\[d(t) = \begin{cases} 55t & \text{for } 0 \leq t \leq 2 \45t + 20 & \text{for } t > 2 \end{cases}\]This notation precisely captures the changeover point at 2 hours, and enables easy adjustments if variables change. For students, seeing the function written this way simplifies understanding the separate components and their conditions.

Precalculus Problem Solving

When solving precalculus problems involving functions and modeled movement, understanding the problem context is crucial. Breaking down the problem into smaller parts makes the process manageable and grants clarity.

• Identify the rates and durations at which each rate applies. Separate the problem into logical segments.
• Translate these segments into mathematical expressions, using the correct units and making sure to account for all variables and assumptions.
• Finally, compile the expressions into a piecewise function if necessary, which allows a comprehensive overview and validation of the solution through simplification.

By following these steps, you develop not only the solution but a deeper understanding of how changes in variables affect the outcome, honing crucial problem-solving skills in the process.