/*! 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 20 The motion of a body falling fro... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 20

The motion of a body falling from rest in a resisting medium is described by the equation $$ \frac{d v}{d t}=A-B v $$ where \(A\) and \(B\) are constants. Then (a) initial acceleration of the body is \(A\) (b) the velocity at which acceleration becomes zero is \(A / B\) (c) the velocity at any time \(t\) is \(\frac{A}{B}\left(1-e^{B t}\right)\) (d) All of the above are wrong

Short Answer

Expert verified
All given options are correct; therefore, (d) is wrong.

Step by step solution

01

Understand the Initial Condition

We are given the differential equation \( \frac{d v}{d t}=A-B v \), which describes the rate of change of velocity \( v \) with time \( t \). The body falls from rest, implying the initial velocity \( v(0) = 0 \).
02

Calculate the Initial Acceleration

The initial acceleration can be found by substituting \( v = 0 \) into the differential equation: \( \frac{d v}{d t} = A - B \times 0 = A \). Thus, the initial acceleration of the body is \( A \).
03

Determine Velocity at Zero Acceleration

The acceleration becomes zero when \( \frac{d v}{d t} = 0 \), implying \( A - Bv = 0 \). Solving for \( v \), we get \( v = \frac{A}{B} \). This is the velocity at which acceleration becomes zero.
04

Solve the Differential Equation for Velocity at Time t

Rearrange the differential equation to separate variables: \( \frac{d v}{A - Bv} = dt \). Integrating both sides, we get \( -\frac{1}{B} \ln |A-Bv| = t+C \). Solving for \( v \) and using the initial condition \( v(0) = 0 \), we obtain \( v(t) = \frac{A}{B} (1 - e^{-Bt}) \).
05

Ascertain the Correct Option

Option (a) states the initial acceleration is \( A \), which matches our findings in Step 2. Option (b) states the velocity at zero acceleration is \( \frac{A}{B} \), which we verified in Step 3. Option (c) provides the velocity at time \( t \), which we derived in Step 4. Hence, all these statements are correct.

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.

Motion in a Resisting Medium
When a body moves through a resisting medium, such as air or water, the resistance from the medium impacts its motion. This resistance often acts as a damping force opposing the motion of the object. In our exercise, the equation \( \frac{d v}{d t}=A-B v \) defines this phenomenon for a body falling from rest.
  • Resisting Medium: The medium exerts a force that depends on the velocity of the falling object. The resisting force is proportional to the velocity, which is captured by the term \(-Bv\) in our equation.
  • Differential Equation: The presence of a differential equation indicates that a function and its derivatives are involved, representing physical quantities like velocity and time.
This kind of equation helps us predict how the body's speed changes over time, making it a key element in describing motion within physics.
Initial Conditions in Differential Equations
Initial conditions are crucial in solving differential equations as they provide the specific information required to find unique solutions. In the context of motion in a resisting medium, we start with the object at rest, which translates to an initial velocity of zero.
  • Importance of Initial Conditions: Initial conditions, like \(v(0) = 0\), allow us to solve the equation appropriately for this particular scenario.
  • Calculating Initial Values: From the differential equation \(\frac{d v}{d t}=A-B v\), substituting \(v=0\) provides immediate acceleration as \(A\).
Knowing the initial velocity helps us plot the motion of the object over time, ensuring we have a complete and accurate picture of the system's dynamics.
Velocity-Time Relationship
The relationship between velocity and time is pivotal in understanding how quickly or slowly an object moves through a resisting medium. For the given differential equation, the velocity as a function of time was derived by integrating the equation.
  • Equation Transformation: By separating variables and integrating, you obtain \( v(t) = \frac{A}{B} (1 - e^{-Bt}) \).
  • Exponential Decay: The term \(e^{-Bt}\) represents an exponential decay, illustrating how the velocity changes over time. This decay is characteristic of motion resistance, where the speed approaches a terminal velocity of \(\frac{A}{B}\).
This relationship provides insights into how the object accelerates initially but gradually reaches a stable velocity as the resisting medium's force balances other forces exerted on the object.

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 particle moving in a straight line covers half the distance with speed of \(3 \mathrm{~m} / \mathrm{s}\). The other half of the distance is covered in two equal time intervals and with speeds of \(4.5 \mathrm{~m} / \mathrm{s}\) and \(7.5 \mathrm{~m} / \mathrm{s}\) respectively. The average speed of the particle during this motion is (a) \(4.0 \mathrm{~m} / \mathrm{s}\) (b) \(5.0 \mathrm{~m} / \mathrm{s}\) (c) \(5.5 \mathrm{~m} / \mathrm{s}\) (d) \(4.8 \mathrm{~m} / \mathrm{s}\)

A particle starts from rest and travels a distance \(s\) with uniform acceleration, then it travels a distance \(2 s\) with uniform speed, finally it travels a distance \(3 s\) with uniform retardation and comes to rest. If the complete motion of the particle in a straight line then the ratio of its average velocity to maximum velocity in (a) \(6 / 7\) (b) \(4 / 5\) (c) \(3 / 5\) (d) \(2 / 5\)

A point initially at rest moves along \(x\)-axis. Its acceleration varies with time as \(a=(6 t+5) \mathrm{m} / \mathrm{s}^{2} .\) If it starts from origin, the distance covered in \(2 \mathrm{~s}\) is (a) \(20 \mathrm{~m}\) (b) \(18 \mathrm{~m}\) (c) \(16 \mathrm{~m}\) (d) \(25 \mathrm{~m}\)

A stone is allowed to fall from the top of a tower \(100 \mathrm{~m}\) high and at the same time another stone is projected vertically upwards from the ground with a velocity of \(254 \mathrm{~ms}^{-1}\). The two stones will meet after \(\begin{array}{llll}\text { (a) } 4 \mathrm{~s} & \text { (b) } 0.4 \mathrm{~s} & \text { (c) } 0.04 \mathrm{~s} & \text { (d) } 40 \mathrm{~s}\end{array}\)

A body is moving along a straight line path with constant velocity. At an instant of time the distance of time the distance travelled by it is \(s\) and its displacement is \(D\), then (a) \(D5\) (c) \(D=s\) (d) \(D \leq s\)
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Physical Quantities And Units

Read Explanation

Energy Physics

Read Explanation

Fields in Physics

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.