/*! 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 43 The aerodynamic resistance to mo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 43

The aerodynamic resistance to motion of a car is nearly proportional to the square of its velocity. Additional frictional resistance is constant, so that the acceleration of the car when coasting may be written \(a=-C_{1}-C_{2} v^{2},\) where \(C_{1}\) and \(C_{2}\) are constants which depend on the mechanical configuration of the car. If the car has an initial velocity \(v_{0}\) when the engine is disengaged, derive an expression for the distance \(D\) required for the car to coast to a stop.

Short Answer

Expert verified
The distance to coast to a stop is \( D = \frac{1}{C_1}\sqrt{\frac{C_1}{C_2}} \tan^{-1}\left( \frac{v_0\sqrt{C_2}}{\sqrt{C_1}} \right) \).

Step by step solution

01

Understand the equation of motion

The given equation is for acceleration: \( a = -C_{1} - C_{2}v^{2} \), where \( a \) is the acceleration of the car, \( C_{1} \) is the constant frictional resistance, \( C_{2}v^2 \) represents aerodynamic resistance, and \( v \) is velocity.
02

Transform the acceleration equation

Using the relationship between acceleration and velocity, \( a = \frac{dv}{dt} \), rewrite the equation as \( \frac{dv}{dt} = -C_{1} - C_{2}v^{2} \). Then, rearrange it to separate variables: \( \frac{dv}{C_{1} + C_{2}v^{2}} = -dt \).
03

Integrate to find time as a function of velocity

Integrate both sides: \( \int \frac{1}{C_{1} + C_{2}v^{2}}\, dv = -\int dt \). Solve the integral using a trigonometric substitution for the left side, which will give a function \( t(v) \).
04

Solve the integral

The integral \( \int \frac{1}{C_{1} + C_{2}v^{2}}\, dv \) can be solved by substituting \( v = \sqrt{\frac{C_1}{C_2}}\tan(\theta) \), resulting in \( \frac{1}{\sqrt{C_1C_2}} \tan^{-1} \left( \frac{v}{\sqrt{\frac{C_1}{C_2}}} \right ) = -t + C \).
05

Simplify and solve for distance

The distance covered is obtained by integrating velocity over time: \( D = \int v dt \). Use \( v = \frac{dv}{a} \) and substitute from earlier steps, finally integrate to find \( D \) in terms of given variables. After simplification, the expression \( D = \frac{1}{C_1}\sqrt{\frac{C_1}{C_2}} \tan^{-1}\left( \frac{v_0\sqrt{C_2}}{\sqrt{C_1}} \right) \) is obtained.

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.

Aerodynamic Resistance
Aerodynamic resistance refers to the forces that oppose a vehicle's motion through the air. These forces are primarily driven by the shape and surface area of the vehicle. This concept is critical because when a car moves, it has to push through the air, creating resistance that increases with speed. In our given formula, aerodynamic resistance is expressed through the term \(C_2v^2\). Here鈥檚 why:
  • The faster the car goes, the larger the force pushing against it.
  • This relationship is quadratic, meaning it increases with the square of the velocity \(v\).
  • This resistance works against the car, slowing it down, hence the negative sign in the equation.
Understanding aerodynamic resistance helps engineers design cars that minimize this drag, improving fuel efficiency and performance.
Coasting Motion Equations
Coasting motion equations describe what happens when a vehicle moves without power from the engine. This usually happens once a car reaches a desired speed and the engine is disengaged.
  • The primary equation involved in our problem is \( a = -C_1 - C_2v^2 \).
  • This equation reflects how different forces slow down the car.
  • \(C_1\) accounts for constant frictional resistance, like the friction between the car鈥檚 tires and the road.

The use of coasting motion equations enables us to predict how long and how far a car can travel when the forces acting against it are only aerodynamic and frictional.
They give us insight into vehicle behaviors like deceleration patterns and stopping distances, critical for safety and efficiency design.
Velocity-Dependent Acceleration
Velocity-dependent acceleration is a fascinating principle where the acceleration of an object is directly linked to its velocity. In this case, you observe it through the equation \(a = -C_1 - C_2v^2\).
  • The equation shows that as velocity \(v\) increases, the acceleration becomes more negative.
  • Some key factors causing this relationship include air drag and friction which increase at higher speeds.
  • This principle is vital as it demonstrates why vehicles slow naturally when there is no ongoing power input, thanks to forces like aerodynamic drag.

Velocity-dependent acceleration indicates that at high speeds, the car will decelerate more quickly until it eventually stops. This mathematical interaction between velocity and acceleration is a cornerstone of road safety analysis and improving vehicle designs for energy efficiency.

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

Consider the polar axis of the earth to be fixed in space and compute the magnitudes of the velocity and acceleration of a point \(P\) on the earth's surface at latitude \(40^{\circ}\) north. The mean diameter of the earth is \(12742 \mathrm{km}\) and its angular velocity is \(0.7292\left(10^{-4}\right) \mathrm{rad} / \mathrm{s}\)

The driver of the truck has an acceleration of \(0.4 g\) as the truck passes over the top \(A\) of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is \(98 \mathrm{m}\) and the center of mass \(G\) of the driver (considered a particle) is \(2 \mathrm{m}\) above the road. Calculate the speed \(v\) of the truck.

A minivan starts from rest on the road whose constant radius of curvature is \(40 \mathrm{m}\) and whose bank angle is \(10^{\circ} .\) The motion occurs in a horizontal plane. If the constant forward acceleration of the minivan is \(1.8 \mathrm{m} / \mathrm{s}^{2},\) determine the magnitude \(a\) of its total acceleration 5 seconds after starting.

The radar tracking antenna oscillates about its vertical axis according to \(\theta=\theta_{0} \cos \omega t,\) where \(\omega\) is the constant circular frequency and \(2 \theta_{0}\) is the double amplitude of oscillation. Simultaneously, the angle of elevation \(\phi\) is increasing at the constant rate \(\dot{\phi}=K .\) Determine the expression for the magnitude \(a\) of the acceleration of the signal horn \((a)\) as it passes position \(A\) and \((b)\) as it passes the top position \(B,\) assuming that \(\theta=0\) at this instant.

In the design of an amusement-park ride, the cars are attached to arms of length \(R\) which are hinged to a central rotating collar which drives the assembly about the vertical axis with a constant angular rate \(\omega=\dot{\theta} .\) The cars rise and fall with the track according to the relation \(z=(h / 2)(1-\cos 2 \theta)\) Find the \(R\) -, \(\theta\) -, and \(\phi\) -components of the velocity \(\mathbf{v}\) of each car as it passes the position \(\theta=\pi / 4\) rad.
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Oscillations

Read Explanation

Astrophysics

Read Explanation

Physical Quantities And Units

Read Explanation

Electrostatics

Read Explanation

Fluids

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.