/*! 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 25 Suppose you read about a new car... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 25

Suppose you read about a new car that can go from 0 to \(100 \mathrm{km} / \mathrm{h}\) in only 2.0 seconds. What is this car's acceleration? a. about \(50 \mathrm{km} / \mathrm{h}\) b. about \(14 \mathrm{m} / \mathrm{s}^{2}\) c. about \(50 \mathrm{km} / \mathrm{s}^{2}\) d. about \(200 \mathrm{km}\) e. about \(0.056 \mathrm{km} / \mathrm{h}^{2}\)

Short Answer

Expert verified
Option b: 14 m/s虏

Step by step solution

01

Convert initial and final speeds to consistent units

The given final speed is 100 km/h. Convert it to meters per second for ease of calculation: \[ 100 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{100000 \text{ m}}{3600 \text{ s}} \approx 27.78 \text{ m/s} \] Since the car starts from rest, the initial speed is 0 m/s.
02

Use the acceleration formula

Acceleration, \( a \), is given by the formula: \[ a = \frac{\text{change in velocity}}{\text{time}} \] In this case, the change in velocity (\( \text{final speed} - \text{initial speed} \)) is \( 27.78 \text{ m/s} \) - \( 0 \text{ m/s} \), and time is 2.0 seconds.
03

Calculate the acceleration

Plug in the values to find the acceleration: \[ a = \frac{27.78 \text{ m/s} - 0 \text{ m/s}}{2.0 \text{ s}} = \frac{27.78 \text{ m/s}}{2.0 \text{ s}} = 13.89 \text{ m/s}^2 \]Rounding to the nearest whole number, the acceleration is approximately 14 \( \text{m/s}^2 \).
04

Choose the correct option

Compare the calculated acceleration to the given options. The closest match is **Option b: 14 m/s虏**.

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Key Concepts

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

physics
Understanding the concept of acceleration is key in physics, especially when studying motion. Acceleration describes how quickly an object speeds up or slows down. It's a vector quantity, meaning it has both a magnitude and a direction. In practical terms, when you press the gas pedal of a car, you're experiencing acceleration. Newton's Second Law of Motion states that the force on an object is equal to its mass times its acceleration: \( F = ma \).Understanding acceleration helps in predicting how quickly an object can move from a state of rest, or how it can reach a specific speed in a given time.
unit conversion
Unit conversion is essential when solving physics problems, as it ensures we are using consistent units for all calculations. In the given exercise, we converted the speed from kilometers per hour (km/h) to meters per second (m/s). This makes calculations easier and more accurate. To convert km/h to m/s:
Multiply by 1000 to convert kilometers to meters.
Then divide by 3600 to convert hours to seconds. This is because there are 1000 meters in a kilometer and 3600 seconds in an hour. So, 100 km/h becomes approximately 27.78 m/s:
\( 100 \times \frac{1000}{3600} = 27.78 \text{ m/s} \).Different units are often used for different types of measurements due to convenience or convention, so mastering unit conversion is a valuable skill.
motion equations
To calculate acceleration, we use one of the basic motion equations in physics:
and \text{Acceleration} \( a = \frac{\text{change in velocity}}{\text{time}} \) Given the final speed (27.78 m/s), initial speed (0 m/s), and time (2.0 s), we substitute these values into the formula: \( a = \frac{27.78 \text{ m/s} - 0 \text{ m/s}}{2.0 \text{ s}} = 13.89 \text{ m/s}^2 \).
This equation simplifies the process of determining how quickly an object's speed is changing. By understanding and applying motion equations, students can solve various real-world problems related to motion, such as calculating distances, speeds, and accelerations.
Recognizing these fundamental principles empowers students to tackle more complex physics problems with confidence.

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Most popular questions from this chapter

Galileo observed that Venus had phases that correlated with its size in his telescope. From this information, you may conclude that Venus a. is the center of the Solar System. b. orbits the Sun. c. orbits Earth. d. orbits the Moon.

During the latter half of the 1 9 th century, a few astronomers thought there might be a planet circling the Sun inside Mercury's orbit. They even gave it a name: Vulcan. We now know that Vulcan does not exist. If a planet with an orbit one-fourth the size of Mercury's actually existed, what would be its orbital period relative to that of Mercury?

Imagine you throw a ball straight up in the air. At the top of its flight, it stops and then falls again. What is the acceleration of the ball at the top of its flight?

Suppose you read in the newspaper that a new planet has been found. Its average speed in orbit is \(33 \mathrm{km} / \mathrm{s}\). When it is closest to its star it moves at \(31 \mathrm{km} / \mathrm{s}\), and when it is farthest from its star it moves at \(35 \mathrm{km} / \mathrm{s}\). This story is in error because a. the average speed is far too fast. b. Kepler's third law says the planet has to sweep out equal areas in equal times, so the speed of the planet cannot change. c. planets stay at a constant distance from their stars; they don't move closer or farther away. d. Kepler's second law says the planet must move fastest when it's closest, not when it is farthest away. e. using these numbers, the square of the orbital period will not be equal to the cube of the semimajor axis.

Suppose you watch a car sliding on ice down a hill. As it slides, it slowly drifts to your right across the road, speeding up in that direction until it bumps into the curb. In this situation, which of the following is more likely, and why? a. The car was already traveling diagonally across the street before it hit the patch of ice. b. The road surface is rounded (or "crowned"), causing the car to be pushed across the street.
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