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Chapter 17: Problem 14

Truck \(\mathrm{X}\) is 13 miles ahead of Truck \(\mathrm{Y}\), which is traveling the same direction along the same route as Truck X. If Truck \(\mathrm{X}\) is traveling at an average speed of 47 miles per hour and Truck \(Y\) is traveling at an average speed of 53 miles per hour, how long will it take Truck Y to overtake and drive 5 miles ahead of Truck X? A 2 hours B 2 hours 20 minutes C 2 hours 30 minutes D 2 hours 45 minutes E 3 hours

Short Answer

Expert verified
3 hours

Step by step solution

01

- Define the variables

Let the time taken for Truck Y to overtake and drive 5 miles ahead of Truck X be \( t \) hours.
02

- Calculate the relative speed

Truck Y is traveling at 53 mph and Truck X is traveling at 47 mph. The relative speed of Truck Y with respect to Truck X is: \[ 53 \text{ mph} - 47 \text{ mph} = 6 \text{ mph} \]
03

- Calculate the distance to overtake

Truck Y needs to cover the 13 miles gap and an additional 5 miles to be ahead by 5 miles. So, the total distance Truck Y needs to cover is: \[ 13 \text{ miles} + 5 \text{ miles} = 18 \text{ miles} \]
04

- Use the formula for time

The time taken is given by distance divided by speed. Using the relative speed, the time taken for Truck Y to overtake and drive 5 miles ahead is: \[ t = \frac{18 \text{ miles}}{6 \text{ mph}} \]
05

- Solve for time

\[ t = 3 \text{ hours} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

relative speed
When two objects move in the same direction, but at different speeds, the concept of relative speed helps us understand how quickly one object is changing their position with respect to the other. In this problem, Truck X and Truck Y are both traveling in the same direction, but Truck Y is faster.
The relative speed is simply the difference between their speeds:
Truck Y's speed - Truck X's speed = 53 mph - 47 mph = 6 mph.
This means that Truck Y is closing the gap between itself and Truck X at a rate of 6 miles per hour. Knowing the relative speed is crucial for calculating how long it will take one object to catch up and surpass another.
distance-time calculations
In problems like this one, we often need to determine how long it takes for one object to catch up to another. This can be solved by understanding the relationship of distance, speed, and time. The basic formula connecting these three is:
\[\text{time} = \frac{\text{distance}}{\text{speed}}\].
In our exercise, Truck Y needs to cover a total additional distance of 18 miles (13 miles to close the gap and 5 more miles to get ahead). We already know Truck Y's relative speed (6 mph). Hence, we use the formula:
\[\text{time} = \frac{18 \text{ miles}}{6 \text{ mph}} = 3 \text{ hours}\].
This tells us that it will take Truck Y 3 hours to not only catch up but also get 5 miles ahead of Truck X.
algebra
Algebra is fundamental in solving many distance-time problems. Setting up the right equations and variables can streamline solving such problems. Let's recap how it鈥檚 used here:
1. Define the variables: Let the time Truck Y needs to overtake and get 5 miles ahead of Truck X be t hours.
2. Use the data given: Truck Y speed: 53 mph, Truck X speed: 47 mph, and initial distance gap: 13 miles + 5 miles ahead.
3. Calculate the relative speed as 6 mph.
4. Use the formula for time based on the total distance and relative speed: \[\text{time} = \frac{18 \text{ miles}}{6 \text{ mph}}\].
By solving for t, which is 3 hours, we get the answer.
Understanding and setting up such equations is key in algebra to solve various real-world problems efficiently.

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Most popular questions from this chapter

A motorcyclist started riding at highway marker \(\mathrm{A},\) drove 120 miles to highway marker \(\mathrm{B}\), and then, without pausing, continued to highway marker \(C,\) where she stopped. The average speed of the motorcyclist, over the course of the entire trip, was 45 miles per hour. If the ride from marker \(A\) to marker \(B\) lasted 3 times as many hours as the rest of the ride, and the distance from marker \(\mathrm{B}\) to marker \(\mathrm{C}\) was half of the distance from marker \(\mathrm{A}\) to marker \(\mathrm{B},\) what was the average speed, in miles per hour, of the motorcyclist while driving from marker \(\mathrm{B}\) to marker \(\mathrm{C} ?\) A 40 B 45 C 50 D 55 E 60

Magnabulk Corp sells boxes holding \(d\) magnets each. The boxes are shipped in crates, each holding \(b\) boxes. What is the price charged per magnet, in cents, if Magnabulk charges \(m\) dollars for each crate? A. $$\frac{100 b d}{m}$$ B. $$\frac{100 m}{b d}$$ C. $$\frac{b d}{100 m}$$ D. $$\frac{m}{b d}$$ E. $$\frac{b d}{m}$$

Did Jon complete a journey of 40 kilometers in less time than it took Ann to complete the same journey? (A) Jon traveled at an average speed of 30 kilometers per hour for the first 10 kilometers and then at an average speed of 15 kilometers per hour for the rest of the journey. (B) Ann traveled at an average speed of 20 kilometers per hour for the entire journey.

A truck driver drove for 2 days. On the second day, he drove 3 hours longer and at an average speed of 15 miles per hour faster than he drove on the first day. If he drove a total of 1,020 miles and spent 21 hours driving during the 2 days, what was his average speed on the first day, in miles per hour? A 25 B 30 C 35 D 40 E 45

A certain account pays 1.5 percent compound interest every 3 months. A person invested an initial amount and did not invest any more money in the account after that. If after exactly 5 years, the amount of money in the account was \(T\) dollars, which of the following is an expression for the original number of dollars invested in the account? A $$(1.015)^{4} T$$ B $$(1.015)^{15} T$$ C $$(1.015)^{20} T$$ D $$\frac{T}{(1.015)^{15}}$$ E $$\frac{T}{(1.015)^{20}}$$
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