/*! 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 4 A car travels up a hill at a con... [FREE SOLUTION] | 91Ó°ÊÓ

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A car travels up a hill at a constant speed of \(40 \mathrm{~km} / \mathrm{h}\) and returns down the hill at a constant speed of \(60 \mathrm{~km} / \mathrm{h}\). Calculate the average speed for the round trip.

Short Answer

Expert verified
The average speed for the round trip is 48 km/h.

Step by step solution

01

Define the problem

To find the average speed for the round trip, we need to consider the total distance traveled and the total time taken.
02

Calculate distance and time for uphill travel

Assume the distance up the hill is \(d\) kilometers. The car travels uphill at a speed of \(40 \mathrm{~km/h}\). The time taken to travel uphill is given by \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{d}{40} \) hours.
03

Calculate distance and time for downhill travel

The distance downhill is the same, \(d\) kilometers. The car travels downhill at a speed of \(60 \mathrm{~km/h}\). The time taken to travel downhill is \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{d}{60} \) hours.
04

Calculate total distance and total time

The total distance for the round trip is \(d + d = 2d\). The total time taken is \(\frac{d}{40} + \frac{d}{60}\) hours.
05

Simplify the expression for total time

Find a common denominator to add the fractions: \(\frac{d}{40} + \frac{d}{60} = \frac{3d}{120} + \frac{2d}{120} = \frac{5d}{120} = \frac{d}{24}\) hours.
06

Calculate the average speed

The average speed is defined as the total distance divided by the total time. Substitute the values: \(\text{Average Speed} = \frac{2d}{\frac{d}{24}} = 2d \times \frac{24}{d} = 48 \mathrm{~km/h}\).

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Key Concepts

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

Constant Speed
When someone refers to constant speed, they mean that an object moves at the same speed over a period of time. It does not slow down or speed up. This is an essential concept in understanding motion, where speed remains uniform throughout the journey. In our car example, as it travels uphill at a constant speed of 40 km/h, it means the car maintains this speed all the way up the hill. Similarly, while coming down the hill, the car maintains a constant speed of 60 km/h.

Constant speed is crucial because it allows us to use simple formulas to calculate time, distance, and speed. The key formula here is:
  • Speed = Distance / Time
This formula can be manipulated to solve for any unknown variable, provided you have the other two. For example, if you have speed and distance, you can find the time by rearranging it to:
  • Time = Distance / Speed
Understanding constant speed allows for straightforward calculations and predictions about motion in uniform conditions.
Distance and Time
Distance and time are two fundamental concepts in physics and everyday life that describe motion. Distance refers to how much ground an object covers during its journey, while time is the duration taken to cover that distance.

In our exercise, the distance is represented as a variable, denoted by "d". The car travels this distance both uphill and downhill. What's unique about this situation is that the distance remains unchanged; it's the same both ways.
The time taken for any trip depends on both the speed of travel and the distance to be covered. Using the formula:
  • Time = Distance / Speed
we can determine the time spent traveling in each direction. For uphill travel, the time is calculated as \(\frac{d}{40}\) hours, and for downhill, it is \(\frac{d}{60}\) hours. Summing up these times gives the total time for the entire trip. By understanding the interplay between distance and time, we can better analyze and solve motion-related problems.
Round Trip
A round trip involves traveling from one point to another and then returning to the original point. It includes both the journey to the destination and the return journey. This concept is crucial because it involves the calculation of total distance and total time, necessary for finding the average speed.

In our exercise, the car embarks on a round trip up and down a hill. The key is to recognize that the distance uphill is equal to the distance downhill, making the total distance for the round trip twice the single path distance, or \(2d\).For the total time, as calculated:
  • Total Time = \(\frac{d}{40} + \frac{d}{60}\)
Finding the average speed for a round trip can sometimes be confusing because it isn't just the average of the two speeds (uphill and downhill). Instead, the average speed is the total distance divided by the total time:
  • Average Speed = Total Distance / Total Time
For this problem, it comes out to be: \(48\mathrm{~km/h}\), illustrating why understanding the full round trip calculation is necessary for accurate results.

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Most popular questions from this chapter

A car moves along an \(x\) axis through a distance of \(900 \mathrm{~m}\), starting at rest (at \(x=0)\) and ending at rest (at \(x=900 \mathrm{~m})\). Through the first \(\frac{1}{4}\) of that distance, its acceleration is \(+2.25 \mathrm{~m} / \mathrm{s}^{2}\). Through the rest of that distance, its acceleration is \(-0.750 \mathrm{~m} / \mathrm{s}^{2}\). What are (a) its travel time through the \(900 \mathrm{~m}\) and \((\mathrm{b})\) its maximum speed? (c) Graph position \(x\), velocity \(v\), and acceleration \(a\) versus time \(t\) for the trip.

A key falls from a bridge that is \(45 \mathrm{~m}\) above the water. It falls directly into a model boat, moving with constant velocity, that is \(12 \mathrm{~m}\) from the point of impact when the key is released. What is the speed of the boat?

You drive on Interstate 10 from San Antonio to Houston, half the time at \(55 \mathrm{~km} / \mathrm{h}\) and the other half at \(90 \mathrm{~km} / \mathrm{h}\). On the way back you travel half the distance at \(55 \mathrm{~km} / \mathrm{h}\) and the other half at \(90 \mathrm{~km} / \mathrm{h}\). What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch \(x\) versus \(t\) for (a), assuming the motion is all in the positive \(x\) direction. Indicate how the average velocity can be found on the sketch.

To test the quality of a tennis ball, you drop it onto the floor from a height of \(4.00 \mathrm{~m}\). It rebounds to a height of \(2.00 \mathrm{~m}\). If the ball is in contact with the floor for \(12.0 \mathrm{~ms}\), (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down?

When startled, an armadillo will leap upward. Suppose it rises \(0.544 \mathrm{~m}\) in the first \(0.200 \mathrm{~s}\). (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of \(0.544 \mathrm{~m}\) ? (c) How much higher does it go?

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