Question

Abdul recently made a 200 mile trip. For the first 30 miles, he traveled at an average speed of 45 miles per hour. His average speed for the next 50 miles was 60 mph. Abdul averaged 50 mph for the final portion of his trip. How long did it take Abdul to complete his journey?

Short answer

It took Abdul 3.9 hours to complete his journey.

Step-by-step solution

Calculate Time for First Segment

For the first 30 miles at an average speed of 45 mph, use the formula for time: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] \[ \text{Time}_1 = \frac{30 \text{ miles}}{45 \text{ mph}} = \frac{2}{3} \text{ hours} \]

Calculate Time for Second Segment

For the next 50 miles at an average speed of 60 mph, the time is: \[ \text{Time}_2 = \frac{50 \text{ miles}}{60 \text{ mph}} = \frac{5}{6} \text{ hours} \]

Determine Distance for Final Segment

The total trip is 200 miles. Subtract the distances of the first and second segments to find the distance for the final segment: \[ 200 - 30 - 50 = 120 \text{ miles} \]

Calculate Time for Final Segment

For the final 120 miles at an average speed of 50 mph, the time is: \[ \text{Time}_3 = \frac{120 \text{ miles}}{50 \text{ mph}} = 2.4 \text{ hours} \]

Sum All Time Segments

Add the time for all three segments to find the total time: \[ \text{Total Time} = \frac{2}{3} + \frac{5}{6} + 2.4 \] Convert all times to a common denominator if needed: \[ \frac{2}{3} = 0.6667 \text{ hours}, \frac{5}{6} = 0.8333 \text{ hours} \] Summing these: \[ 0.6667 + 0.8333 + 2.4 = 3.9 \text{ hours} \]

Key concepts

Average Speed Calculation

Understanding how to calculate average speed is crucial for solving time-distance problems. Average speed is the total distance traveled divided by the total time taken. The formula is given by:

\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]

For instance, in the problem involving Abdul's trip, you can determine the average speed for each segment of the journey by using this formula. You need the distances and speeds for each segment:

- First 30 miles at 45 mph
- Next 50 miles at 60 mph
- Final 120 miles at 50 mph

By calculating the time taken for each part separately and adding them, you can find the total time for the journey. Finally, if you need the overall average speed, use the total distance (200 miles) and the total time (3.9 hours) to get the average speed over the entire trip.

Distance-Time Relationship

The relationship between distance and time is fundamental in solving these problems. The basic formula connecting distance \(d\), speed \(s\), and time \(t\) is:

\[ d = s \times t \]

This equation tells us how distance, speed, and time interact. If you know any two of these variables, you can find the third. For example:

- To find the time taken \(t\) for Abdul’s first segment: \[ t = \frac{d}{s} = \frac{30 \text{ miles}}{45 \text{ mph}} = \frac{2}{3} \text{ hours} \]
- Likewise, for the second segment: \[ t = \frac{d}{s} = \frac{50 \text{ miles}}{60 \text{ mph}} = \frac{5}{6} \text{ hours} \]

By applying this formula to each segment of the journey and then summing the times, you can solve for the total time taken.

Algebraic Problem-Solving

Algebra helps us solve for unknowns in distance-time problems. It provides a structured way to break down each segment and solve them step-by-step. Here’s how you can approach it:

1. **Segment the Problem**: Split the journey into parts, each with known distance and speed. Break Abdul's 200-mile trip into 30 miles, 50 miles, and 120 miles segments.

2. **Use the Formula**: Apply the formula \[ t = \frac{d}{s} \] to find the time for each segment.
- First Segment: \[ t_1 = \frac{30}{45} = \frac{2}{3} \]
- Second Segment: \[ t_2 = \frac{50}{60} = \frac{5}{6} \]
- Third Segment: \[ t_3 = \frac{120}{50} = 2.4 \]

3. **Sum the Times**: Add the times for all segments to find the total time.
\[ \text{Total Time} = t_1 + t_2 + t_3 = \frac{2}{3} + \frac{5}{6} + 2.4 = 3.9 \text{ hours} \]

With algebra, you keep track of each component in the problem, ensuring that you solve it systematically and accurately.