/*! 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 21 For a standard production car, t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 21

For a standard production car, the highest road-tested acceleration ever reported occurred in \(1993,\) when a Ford \(\mathrm{RS} 200\) Evolution went from zero to \(26.8 \mathrm{m} / \mathrm{s}(60 \mathrm{mi} / \mathrm{h})\) in \(3.275 \mathrm{s} .\) Find the magnitude of the car's acceleration.

Short Answer

Expert verified
The car's acceleration is approximately 8.19 m/s虏.

Step by step solution

01

Understand the Formula

To find acceleration, we use the formula \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time taken for that change.
02

Identify the Variables

The initial velocity (\( v_0 \)) is 0 m/s. The final velocity (\( v \)) is 26.8 m/s. The time interval (\( \Delta t \)) is 3.275 seconds.
03

Calculate the Change in Velocity

Since the car starts from rest, the change in velocity \( \Delta v \) is equal to the final velocity: \( \Delta v = v - v_0 = 26.8 \text{ m/s} - 0 \text{ m/s} = 26.8 \text{ m/s} \).
04

Apply the Formula

Plug the values into the acceleration formula: \( a = \frac{26.8 \text{ m/s}}{3.275 \text{ s}} \).
05

Perform the Calculation

Calculate \( a = \frac{26.8}{3.275} = 8.186 \text{ m/s}^2 \). Therefore, the magnitude of the acceleration is approximately 8.19 \( \text{ m/s}^2 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the causes of this motion. It focuses primarily on answering how objects move, rather than why they move.
In kinematics, we use mathematical equations to describe the motion of objects. These equations relate various physical quantities such as displacement, velocity, and acceleration.
For example, when considering a car accelerating on a road, kinematics helps us understand the car's change from a stationary position to a moving state.
  • Displacement describes the car鈥檚 change in position.
  • Velocity gives us the rate of change of position over time.
  • Acceleration tells us how quickly the velocity changes.
Understanding these concepts is crucial for solving physics problems related to motion.
Velocity Change
Velocity change, often represented as \( \Delta v \), is a fundamental concept in kinematics. It describes how the speed or velocity of an object varies over a period of time. To determine the change in velocity, you subtract the initial velocity from the final velocity:
\[ \Delta v = v - v_0 \]For instance, suppose a car accelerates from a stop (0 m/s) to a speed of 26.8 m/s. The change in velocity would be calculated as:
  • Initial velocity \( v_0 = 0 \text{ m/s} \)
  • Final velocity \( v = 26.8 \text{ m/s} \)
  • Change in velocity \( \Delta v = 26.8 \text{ m/s} - 0 \text{ m/s} = 26.8 \text{ m/s} \)
Understanding how velocity changes are essential for determining acceleration and analyzing motion in physics.
Acceleration Formula
The acceleration of an object is a measure of how quickly its velocity changes with time. It is calculated using the formula:
\[ a = \frac{\Delta v}{\Delta t} \]where \( a \) is acceleration, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the time interval over which the change occurs.
This formula tells us that acceleration is directly proportional to the change in velocity and inversely proportional to the time over which the change happens.
  • If \( \Delta v \) is large, and \( \Delta t \) is small, the acceleration is high, indicating a rapid increase in speed.
  • If \( \Delta v \) is small, the acceleration is low, showing a gradual speed increase.
Understanding the acceleration formula is crucial, as it enables us to quantify how the speed of an object changes over time.
Time Interval
In physics, the term "time interval" refers to the duration over which a certain change, like velocity change, occurs. It鈥檚 denoted as \( \Delta t \) and is a key component in calculating acceleration. The time interval helps us understand how quickly or slowly a change takes place.
In our car example, the vehicle goes from a complete stop to 26.8 m/s in 3.275 seconds. Thus, \( \Delta t \) is 3.275 seconds. This period is essential for calculating the acceleration as:
  • Shorter time intervals often mean higher acceleration if the velocity change is constant.
  • Longer time intervals indicate slower acceleration rate for the same velocity change.
By considering the time interval, we get a better understanding of the dynamics of motion and how different factors interact to produce changes in velocity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board ), assuming that the downward direction is chosen as the negative direction. \(J\) (b) What is the highest point he reaches above the water?

A car is traveling along a straight road at a velocity of \(+36.0 \mathrm{m} / \mathrm{s}\) when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is \(\bar{a}_{1} .\) For the next six seconds the car slows down further, and its average acceleration is \(\bar{a}_{2} .\) The velocity of the car at the end of the eighteen-second period is \(+28.0 \mathrm{m} / \mathrm{s}\). The ratio of the average acceleration values is \(\bar{a}_{1} / \bar{a}_{2}=1.50 .\) Find the velocity of the car at the end of the initial twelve-second interval.

A locomotive is accelerating at \(1.6 \mathrm{m} / \mathrm{s}^{2} .\) It passes through a \(20.0-\mathrm{m}-\) wide crossing in a time of \(2.4 \mathrm{s} .\) After the locomotive leaves the crossing, how much time is required until its speed reaches \(32 \mathrm{m} / \mathrm{s} ?\)

A skydiver is falling straight down, along the negative \(y\) direction. (a) During the initial part of the fall, her speed increases from 16 to \(28 \mathrm{m} / \mathrm{s}\) in \(1.5 \mathrm{s},\) as in part \(a\) of the figure. (b) Later, her parachute opens, and her speed decreases from 48 to \(26 \mathrm{m} / \mathrm{s}\) in \(11 \mathrm{s},\) as in part \(b\) of the drawing. Concepts: (i) Is her average acceleration positive or negative when her speed is increasing in part \(a\) of the figure? (ii) Is her average acceleration positive or negative when her speed is decreasing in part \(b\) of the figure? Calculations: In both instances (parts \(a\) and \(b\) of the figure) determine the magnitude and direction of her average acceleration.

A ball is thrown straight upward and rises to a maximum height of \(16 \mathrm{m}\) above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fields in Physics

Read Explanation

Modern Physics

Read Explanation

Physics of Motion

Read Explanation

Oscillations

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.