Problem 38 - Elementary Algebra

Question

Dawn starts on a 58-mile trip on her moped at 20 miles per hour. After a while the motor stops, and she pedals the remainder of the trip at 12 miles per hour. The entire trip takes \(3 \frac{1}{2}\) hours. How far had Dawn traveled when the motor on the moped quit running?

Step-by-Step Solution

Answer

Dawn traveled 40 miles when the motor quit running.

Step 1: Define Variables

Let \(d\) be the distance Dawn traveled with the motor running, and \(58 - d\) be the distance pedaled. Let \(t_1\) and \(t_2\) be the time spent during motor travel and pedaling, respectively.

Step 2: Write Equations for Time

The time Dawn traveled using the motor, \(t_1\), is \(\frac{d}{20}\), and the time pedaling, \(t_2\), is \(\frac{58-d}{12}\). The total trip time is \(3.5\) hours, so the equation is:\[t_1 + t_2 = 3.5\]Substitute the expressions for \(t_1\) and \(t_2\) into this equation:\[\frac{d}{20} + \frac{58-d}{12} = 3.5\]

Step 3: Simplify and Solve the Equation

To solve \(\frac{d}{20} + \frac{58-d}{12} = 3.5\), first find a common denominator, which is 60:\[\frac{3d}{60} + \frac{5(58-d)}{60} = 3.5\]Simplifying further:\[3d + 290 - 5d = 210\]Combine like terms:\[-2d + 290 = 210\]Subtract 290 from both sides:\[-2d = -80\]Divide each side by -2:\[d = 40\]

Step 4: Verify the Solution

Substitute \(d = 40\) back into the time equations to ensure the total time is correct.\[t_1 = \frac{40}{20} = 2\text{ hours}\]\[t_2 = \frac{18}{12} = 1.5\text{ hours}\]Sum of times is \(2 + 1.5 = 3.5\) hours, confirming the solution is accurate.

Key Concepts

Rate and time calculations

Understanding the relationship between rate, time, and distance can help solve many real-world travel problems. The basic principle is captured in the formula: distance = rate × time. Knowing any two of these variables allows us to solve for the third.

For instance, when Dawn started her trip using the moped motor, she traveled at a rate of 20 miles per hour. If we denote the distance traveled with the motor as \(d\), then the time \(t_1\) it took can be expressed using the formula:

• \(t_1 = \frac{d}{20}\)

Later, when the motor stopped, Dawn switched to pedaling at 12 miles per hour for the remaining distance \(58 - d\). The time \(t_2\) for this second part of the trip is then:

• \(t_2 = \frac{58-d}{12}\)

These separate segments of the trip highlight how rate and time calculations help break down complex travel scenarios into more manageable parts.

Solving equations

Once the problem is set up with the right equations, solving them becomes straightforward. We start by combining the expressions for time from the two segments of Dawn's trip to match the total travel time.

The equation \(\frac{d}{20} + \frac{58-d}{12} = 3.5\) reflects the sum of the time for traveling with the motor and pedaling. The first step is to find a common denominator, which makes comparisons easier by allowing for combining terms.

• Rewriting each term with the common denominator 60: \(\frac{3d}{60} + \frac{5(58-d)}{60} = 3.5\)
• Simplifying gives \(3d + 290 - 5d = 210\)
• Combining like terms leads to \(-2d + 290 = 210\)

From here, we isolate \(d\) by subtracting 290 from both sides and dividing by -2 to yield \(d = 40\). Solving equations like this helps find unknown values systematically by manipulating algebraic expressions.

Verification of solutions

Verifying the solution is a crucial final step to ensure the calculated result is correct. Once we determine \(d = 40\), substituting this back into the original equations checks if the conditions align with the problem's given constraints.

Substituting \(d = 40\):

• Time with the motor: \(t_1 = \frac{40}{20} = 2\) hours
• Time pedaling: \(t_2 = \frac{18}{12} = 1.5\) hours

Adding these times, we get \(2 + 1.5 = 3.5\) hours, which matches the total given time for the trip. Ensuring both the individual components and the total time agree confirms that the solution is accurate, preventing errors and building confidence in the steps taken.