/*! 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 1 The acceleration of a Maserati i... [FREE SOLUTION] | 91Ó°ÊÓ

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The acceleration of a Maserati is proportional to the difference between \(250 \mathrm{~km} / \mathrm{h}\) and the velocity of this sports car. If this machine can accelerate from rest to \(100 \mathrm{~km} / \mathrm{h}\) in \(10 \mathrm{~s}\), how long will it take for the car to accelerate from rest to \(200 \mathrm{~km} / \mathrm{h}\) ?

Short Answer

Expert verified
The car takes approximately 26.444 seconds to accelerate from rest to 200 km/h.

Step by step solution

01

Understanding the Problem

We are given that the acceleration of the Maserati is proportional to the difference between 250 km/h and its velocity. We need to find how long it takes for the Maserati to go from rest (0 km/h) to 200 km/h.
02

Formulate the Differential Equation

Let the velocity at time \( t \) be \( v(t) \). The acceleration \( a(t) = \frac{dv}{dt} \) is proportional to \( 250 - v(t) \). Thus, \( \frac{dv}{dt} = k (250 - v(t)) \) for some proportionality constant \( k \).
03

Solve the Differential Equation for v(t)

Separate variables: \( \frac{1}{250 - v} dv = k \, dt \). Integrate both sides: \( -\ln|250 - v| = kt + C \). Solve for \( v(t) \): \( 250 - v = Ce^{-kt} \), \( v = 250 - Ce^{-kt} \).
04

Determine the Constant C Using Initial Conditions

Initially, \( v(0) = 0 \). Substitute into the expression: \( 0 = 250 - C \), so \( C = 250 \). Thus, \( v(t) = 250(1 - e^{-kt}) \).
05

Calculate k Using Given Data

Use the condition that \( v(10) = 100 \) km/h. Substitute \( t = 10 \) and \( v(t) = 100 \) into \( v(t) = 250(1 - e^{-kt}) \): \( 100 = 250(1 - e^{-10k}) \). Solve for \( k \): \( e^{-10k} = 0.6 \), giving \( k = -\frac{\ln(0.6)}{10} \).
06

Find Time to Reach 200 km/h

Set \( v(t) = 200 \) and solve \( 200 = 250(1 - e^{-kt}) \). This gives \( e^{-kt} = 0.2 \). Use \( k = -\frac{\ln(0.6)}{10} \) to solve for \( t \). Then \( -kt = \ln(0.2) \), so \( t = \frac{\ln(0.2)}{-\frac{\ln(0.6)}{10}} \).

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Key Concepts

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

Initial Value Problem
An initial value problem in differential equations is one where we need to find the specific solution that not only satisfies a given differential equation but also fulfills an initial condition. This initial condition often represents a starting scenario or an initial moment in time. For example, if we know the velocity of our Maserati at the start, we utilize this information as our initial condition to find the unique solution to our problem.
In our exercise, we begin by letting the velocity, denoted by \( v(t) \), describe the state of the car at any time \( t \). Initially, at \( t = 0 \), we know that the car is at rest, so \( v(0) = 0 \). This starting condition allows us to compute the constant \( C \) in our solution, providing a complete and specific behavior model for the car's acceleration behavior.
  • It's like setting up a specific scenario we want to study.
  • This initial value lets us refine and pinpoint our solution precisely.
Exponential Growth and Decay
Exponential growth and decay events are characterized by their change rate being proportional to their current size or state. Generally, exponential decay occurs in scenarios where items decrease rapidly at first, and then more slowly over time. This behavior is highly relevant when applied to our Maserati's acceleration.
The mathematical representation involves an exponential term. In this case, \( v(t) = 250(1 - e^{-kt}) \) shows how velocity changes over time. The negative exponent indicates a decay since the velocity approaches a constant limit (here, 250 km/h), mimicking the car reaching its top speed potential.
This is quite common in natural processes, such as cooling rates and radioactive decay.
  • Shows how quickly or slowly something changes over time.
  • In our problem, depicts how the car quickly reaches closer to its maximum speed limit.
Velocity and Acceleration
Velocity and acceleration are central concepts in kinematics. Velocity refers to how fast an object moves and its direction, while acceleration describes how fast the velocity changes over time. The relationship between them is key to solving our exercise.
From the differential equation, acceleration is represented as \( a(t) = \frac{dv}{dt} = k(250 - v(t)) \). This means the acceleration of the car depends on the gap between current velocity and a threshold speed, here 250 km/h. When the velocity is lower, acceleration is higher, encouraging the car to speed up faster.
Over time, as the car moves closer to its threshold, the increase in velocity slows.
  • Acceleration tells us about the changing pace of speeding up.
  • This interplay helps us predict when the car will reach desired speeds.
Proportional Relationships
Proportional relationships signify a consistent rate or ratio between two quantities. In many real-world applications, especially in physics, understanding these relationships can simplify complex problems.
In the Maserati exercise, the acceleration is proportionally related to the difference between the velocity and a certain maximum limit (250 km/h). This setup makes it easy to predict changes in acceleration as the car's speed rises. It implies that as one quantity increases (velocity), the amount by which another (acceleration) changes can be directly described by this established ratio or function.
Such relationships make modeling real-world dynamics more straightforward and understandable.
  • Proportionality makes complex interactions simpler and predictable.
  • Helps in making sense of the physics of motion.

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Most popular questions from this chapter

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=e^{-y}, y(0)=0 ; y(x)=\ln (x+1) $$

An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the threedecimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. $$ y^{\prime}=y-x-1, y(0)=1 ; y(x)=2+x-e^{x} $$

Separate variables and use partial fractions to solve the initial value problems in Problems \(1-8 .\) Use either the exact solution or a computer- generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. $$ \frac{d x}{d t}=7 x(x-13), x(0)=17 $$

(a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity \(v_{0}\) less than escape velocity \(\sqrt{2 G M / R}\), then the maximum distance from the center of the earth attained by the projectile is $$ r_{\max }=\frac{2 G M R}{2 G M-R v_{0}^{2}} $$ where \(M\) and \(R\) are the mass and radius of the earth, respectively. (b) With what initial velocity \(v_{0}\) must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth, expressed in terms of earth radii, attained by a projectile launched from the surface of the earth with \(90 \%\) of escape velocity.

Suppose that a community contains 15,000 people who are susceptible to Michaud's syndrome, a contagious disease. At the time \(t=0\) the number \(N(t)\) of people who have developed Michaud's syndrome is 5000 and is increasing at the rate of 500 per day. Assume that \(N^{\prime}(t)\) is proportional to the product of the numbers of those who have caught the disease and of those who have not. How long will it take for another 5000 people to develop Michaud's syndrome?

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