Problem 28 - College Algebra and Calculus: An Applied Approach

Question

Traveled by a Car You are driving at a constant speed. At \(4: 30\) P.M., you drive by a sign that gives the distance to Montgomery, Alabama as 84 miles. At 4:59 P.M., you drive by another sign that gives the distance to Montgomery as 56 miles. (a) Write a linear equation that gives your distance from Montgomery in terms of time \(t\). (Let \(t=0\) represent \(4: 30 \mathrm{P.M}\). and let \(t\) be measured in minutes.) (b) Use the equation in part (a) to find the time when you will reach Montgomery.

Step-by-Step Solution

Answer

The linear equation representing the distance from Montgomery as a function of time is \(d = \frac{28}{29}t + 84\). Using our equation, it's calculated that the driver will reach Montgomery at 5:54 P.M.

Step 1: Determine the Rate of Travel

You are travelling a distance of 84 - 56 = 28 miles in a time of 4:59 P.M. - 4:30 P.M. = 29 minutes. Therefore, the rate is \(rate = \frac{distance}{time} = \frac{28 miles}{29 minutes}\) miles per minute.

Step 2: Write the Linear Equation

The equation will be in the form \(d = mt + b\), where \(d\) is the distance from Montgomery, \(m\) is the rate of travel, \(t\) is the time and \(b\) is the y-intercept. Since you know that at \(t = 0\), \(d = 84 miles\), the y-intercept \(b = 84\). Therefore, the equation becomes \(d = \frac{28}{29}t + 84\).

Step 3: Solve the Equation for the Time

To find the time when you will reach Montgomery, you need to set \(d = 0\) in the equation and solve for \(t\). This gives the equation \(\frac{28}{29}t + 84 = 0\), which simplifies to \(t = -29 \times \frac{84}{28}\). Solving this equation gives \(t = -84\) minutes. However, time cannot be negative. Therefore, this result means you will reach Montgomery 84 minutes after your initial time of 4:30 P.M.

Key Concepts

Rate of Change

When moving at a constant speed, the rate of change is a key concept in understanding motion and can be simply described as how fast something moves over a particular time span. In this exercise, you're given the task of calculating the rate of travel using the information about two distances and the time taken in between these points. The rate of change formula is given by:

• Rate = \( \frac{\text{Distance Change}}{\text{Time Change}} \)

Here, you travel 28 miles in 29 minutes, so the rate is \( \frac{28}{29} \) miles per minute. This rate tells you how many miles you cover every minute. Understanding the rate of change is crucial because it serves as the "slope" in the linear equation that represents the travel. Whenever you encounter problems involving movement, always look out for changes over time to establish this rate.

Distance and Time Problems

Distance and time problems often involve determining how long it will take to reach a destination or finding out a point of arrival given certain conditions. In such problems, you usually have a starting distance and then determine how this distance decreases (or increases) with time as you move.Using the rate calculated from the earlier section, you can find out at any given time how far you are from the destination. Initially, at \( t = 0 \), you are 84 miles away from Montgomery. As time progresses, this distance decreases according to the rate we've calculated.In this exercise, the question focuses on how distance changes as you move closer to the target, illustrating the concept of approaching a destination over time. By understanding distance and time problems, you can apply these principles to various scenarios involving motion and time calculations.

Algebraic Representation

Algebraic representation plays a vital role in translating word problems into mathematical equations that you can solve. In this exercise, the information given is used to form a linear equation to model the situation. The general form of your linear equation is:

• \( d = mt + b \)

where:

• \( d \) is the distance from Montgomery
• \( m \) is the rate of change (your speed)
• \( t \) is the time in minutes after 4:30 P.M.
• \( b \) is the initial distance from Montgomery, which is 84 miles
  

So, your equation becomes \( d = \frac{28}{29} t + 84 \). By setting \( d = 0 \), indicating when you reach Montgomery, you can solve for \( t \) to find how long it takes to arrive. Algebraic representation enables you to easily manipulate and solve the problem, providing a clear pathway from the problem description to the answer. It is a fundamental tool for understanding and solving problems involving relationships between variable quantities.