Problem 67 - Algebra for College Students

Question

Mary drove from Clarksville to Leesville at 45 miles per hour (mph). At Leesville she discovered that she had forgotten her purse. She immediately returned to Clarksville at 55 mph. What was her average speed for the entire trip? (The answer is not 50 mph.)

Step-by-Step Solution

Answer

Mary's average speed for the entire trip was 49.5 mph.

Step 1 - Define Variables

Let the distance between Clarksville and Leesville be denoted by d miles. She travels to Leesville at 45 mph and returns at 55 mph.

Step 2 - Calculate Time for Each Leg of the Trip

Calculate the time to travel to Leesville: \( t_1 = \frac{d}{45} \). Calculate the time to return to Clarksville: \( t_2 = \frac{d}{55} \).

Step 3 - Calculate Total Time

Add the times for each leg of the trip: \( t_{\text{total}} = t_1 + t_2 = \frac{d}{45} + \frac{d}{55} \).

Step 4 - Find Common Denominator

To add the fractions, find a common denominator for 45 and 55, which is 495: \( t_{\text{total}} = \frac{55d + 45d}{45 \times 55} = \frac{100d}{2475} = \frac{20d}{495} \).

Step 5 - Calculate Average Speed

The total distance traveled is \( 2d \). The total time taken is \( \frac{20d}{495} \). The average speed is calculated as follows: \[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{\frac{20d}{495}} = \frac{2d \times 495}{20d} = \frac{990}{20} = 49.5 \text{ mph}. \]

Key Concepts

Distance

Understanding the concept of distance is fundamental in solving average speed problems. In this exercise, distance represents how far Mary travels from Clarksville to Leesville and then back from Leesville to Clarksville. Distance is typically measured in units such as miles, kilometers, or meters. Here, we denote the distance between Clarksville and Leesville as \( d \) miles. When Mary drives from one town to the other, that distance \( d \) is covered twice, once each way.
To find out the total distance, you simply add the distance of each leg of the trip:

• Distance to Leesville: \( d \)
• Distance back to Clarksville: \( d \)
• Total distance: \( 2d \)

In this problem, Mary's total distance traveled is \( 2d \). This is crucial for calculating her average speed.

Speed

Speed is the rate at which an object covers a distance. It is calculated as the distance traveled divided by the time taken to travel that distance.
The formula for speed is:\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
In this exercise, Mary travels to Leesville at a speed of 45 mph and returns to Clarksville at a speed of 55 mph. These speeds determine how quickly she completes each leg of her journey.
It's important to remember that different speeds mean different times for the same distance. Since Mary travels faster on her way back, the time will be shorter compared to going to Leesville.

Time

Time is a critical component when calculating speeds and averages for trips. The time taken to cover a certain distance at a given speed is found using:\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
In this situation, we need to calculate the time for both legs of Mary's trip. To achieve that, we use the given speeds:

• Time to Leesville: \( t_1 = \frac{d}{45} \)
• Time back to Clarksville: \( t_2 = \frac{d}{55} \)

To find the total time, we simply add the two times together:
\[ t_{\text{total}} = t_1 + t_2 = \frac{d}{45} + \frac{d}{55} \]
Next, we find a common denominator to add these fractions. In this case, it's 495:
\[ t_{\text{total}} = \frac{55d + 45d}{45 \times 55} = \frac{100d}{2475} = \frac{20d}{495} \]
Understanding the total time and the total distance (as covered in the previous sections) allows us to calculate the average speed for the entire trip.