/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A student drives the 100 -mi tri... [FREE SOLUTION] | 91Ó°ÊÓ

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A student drives the 100 -mi trip back to campus after spring break and travels with an average speed of \(52 \mathrm{mi} / \mathrm{h}\) for 1 hour and 30 minutes for the first part of the trip. (a) What distance was traveled during this time? (b) Traffic gets heavier, and the last part of the trip takes another half- hour. What was the average speed during this leg of the trip? (c) Find the average speed for the total trip.

Short Answer

Expert verified
(a) 78 miles (b) 44 mi/h (c) 50 mi/h

Step by step solution

01

Calculate Distance for First Part

To find the distance traveled in the first part of the trip, we use the formula for distance: \( \text{Distance} = \text{Speed} \times \text{Time} \). The student travels at an average speed of \(52 \, \text{mi/h}\) for \(1\) hour and \(30\) minutes. Convert the time into hours: \(1 \text{ hour} + 0.5 \text{ hour} = 1.5 \text{ hours}\). Now calculate the distance: \[ \text{Distance} = 52 \, \text{mi/h} \times 1.5 \, \text{h} = 78 \, \text{miles} \].
02

Calculate Remaining Distance

After the first part of 78 miles, the total trip is 100 miles. So, the remaining distance is \[ 100 \, \text{miles} - 78 \, \text{miles} = 22 \, \text{miles} \].
03

Calculate Average Speed for Last Leg

The last part of the trip takes \(0.5\) hours (30 minutes). We use the formula for speed: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \). Thus, the average speed during this leg is \[ \text{Speed} = \frac{22 \, \text{miles}}{0.5 \, \text{hours}} = 44 \, \text{mi/h} \].
04

Calculate Average Speed for Total Trip

To find the total average speed, use \( \text{Total Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \). The total distance of the trip is 100 miles. The total time is \(1.5 + 0.5 = 2\) hours. Therefore, the average speed for the total trip is \[ \text{Total Average Speed} = \frac{100 \, \text{miles}}{2 \, \text{hours}} = 50 \, \text{mi/h} \].

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Key Concepts

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

Average Speed Calculation
Understanding how to calculate average speed is crucial for solving distance and speed problems effectively. Average speed can be thought of as the total distance traveled divided by the total time taken to cover that distance. This is expressed with the formula: \[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \] To illustrate this concept, let's consider the problem scenario. During a total 100-mile journey, the student spent 2 hours on the road. We calculate the student's average speed by dividing the total distance of 100 miles by the total time of 2 hours: \[ \text{Average Speed} = \frac{100 \, \text{miles}}{2 \, \text{hours}} = 50 \, \text{mi/h} \] This means that, on average, the student covered 50 miles each hour during the trip. Calculating average speed this way helps simplify complex journeys into a single rate that represents the overall trip.
Distance Calculation
Distance calculation involves using known speed and time values to find out how far an object or person has traveled. The key formula used here is: \[ \text{Distance} = \text{Speed} \times \text{Time} \] In the problem we examined, the student drove for a duration of 1 hour and 30 minutes at a speed of 52 mi/h for the first part of the trip. There are two steps to solve for distance: - **Convert the time:** Convert 1 hour and 30 minutes into hours. Since 30 minutes is half an hour, this totals 1.5 hours. - **Apply the formula:** Multiply the speed by the time: \[ \text{Distance} = 52 \, \text{mi/h} \times 1.5 \, \text{h} = 78 \, \text{miles} \] The car covered a distance of 78 miles in the first segment of the journey. Knowing how to calculate distance helps in planning travel and estimating arrival times.
Time Conversion
Time conversion is a vital skill in solving distance and speed problems. Often, problems will provide time in mixed units, such as hours and minutes, or require you to convert these into a consistent unit like hours. In the given travel problem, time needed to be converted from 1 hour and 30 minutes to 1.5 hours to seamlessly perform calculations. Here's how time conversion typically works: - **Understanding units:** 60 minutes equals 1 hour. Therefore, to convert minutes to hours, divide the number of minutes by 60. - **Application:** For instance, 30 minutes divided by 60 converts to 0.5 hours. Adding this to 1 hour gives 1.5 hours in total. Such conversions ensure accuracy in computations and make applying formulas more straightforward. Converting time into a single unit simplifies the equations and helps avoid mistakes.

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Most popular questions from this chapter

At a track meet, a runner runs the 100 -m dash in 11.5 s. What was the runner's average speed?

A car travels on a straight, level road. (a) Starting from rest, the car is going \(44 \mathrm{ft} / \mathrm{s}(30 \mathrm{mi} / \mathrm{h})\) at the end of \(5.0 \mathrm{s}\). What is the car's average acceleration in ft/s \(^{2} ?\) (b) In 4.0 more seconds, the car is going \(88 \mathrm{ft} / \mathrm{s}\) ( \(60 \mathrm{mi} / \mathrm{h}\) ). What is the car's average acceleration for this time period? (c) The car then slows to \(66 \mathrm{ft} / \mathrm{s}\) ( \(45 \mathrm{mi} / \mathrm{h}\) ) in 3.0 s. What is the average acceleration for this time period? (d) What is the overall average acceleration for the total time? (Note these convenient British unit conversions: \(60 \mathrm{mi} / \mathrm{h}=88 \mathrm{ft} / \mathrm{s}, 45 \mathrm{mi} / \mathrm{h}=66 \mathrm{ft} / \mathrm{s},\) and \(30 \mathrm{mi} / \mathrm{h}=44 \mathrm{ft} / \mathrm{s} . \mathrm{J}\)

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A race car goes around a circular, level track with a diameter of \(1.00 \mathrm{km}\) at a constant speed of 90.0 km/h. What is the car's centripetal acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) ?

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