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Find the equilibrium solution to \(d v / d t=\) \(-g-(c / m) v\). What is the limiting velocity?

Short Answer

Expert verified
The limiting velocity is \( v_{lim} = -\frac{mg}{c} \).

Step by step solution

01

- Understand the given equation

The differential equation given is \( \frac{dv}{dt} = -g - \frac{c}{m}v \). This represents the rate of change of velocity over time.
02

- Set the equilibrium condition

At equilibrium, the velocity does not change with time. Therefore, set \( \frac{dv}{dt} = 0 \). Substitute this into the equation to get: \ 0 = -g - \frac{c}{m}v \.
03

- Solve for the equilibrium velocity

Rearrange the equation to solve for \( v \). Add \( \frac{c}{m}v \) to both sides: \ \frac{c}{m}v = -g \. Solve for \( v \): \ v = -\frac{mg}{c} \.
04

- Determine the limiting velocity

Since the equilibrium velocity represents the limiting velocity, the limiting velocity \( v_{lim} \) is \ v_{lim} = -\frac{mg}{c} \.

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Key Concepts

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

Rate of change
Understanding the rate of change is essential when dealing with differential equations. Think of the rate of change as how fast something is moving or varying over time. In this context, we're examining the velocity of an object and how it changes. The differential equation given is: \( \frac{dv}{dt} = -g - \frac{c}{m}v \). Here, \( \frac{dv}{dt} \) represents the rate at which the velocity \( v \) changes over time. Understanding this rate is crucial as it helps us determine how quickly the object's velocity is decreasing due to gravity (indicated by -g) and a resistive force proportional to velocity (indicated by \( -\frac{c}{m}v \)).

The negative signs indicate that both gravity and resistance are acting in a direction opposite to the object's motion. Summarizing it:
  • -g: the constant rate due to gravity.
  • -\frac{c}{m}v: the rate of change due to resistive force (like friction or air resistance).
The end goal is to understand how these factors balance out to find the equilibrium condition.
Equilibrium condition
The equilibrium condition is reached when the rate of change of velocity becomes zero. In simpler terms, this means the object has reached a state where its velocity does not change anymore. For any differential equation involving time, an equilibrium condition helps determine a stable solution.

For the given differential equation \( \frac{dv}{dt} = -g - \frac{c}{m}v \), the equilibrium condition can be found by setting the rate of change \( \frac{dv}{dt} \) to zero. This is because, at equilibrium, the velocity is no longer changing:
  • \( 0 = -g - \frac{c}{m}v \).
Solving for the equilibrium velocity \( v \), you'll get:

\ \frac{c}{m}v = -g \
\ v = -\frac{mg}{c} \

This value of \( v \) tells us the steady-state (or equilibrium) velocity of the object when the forces of gravity and resistance have balanced out.
Limiting velocity
The limiting velocity is essentially the equilibrium velocity in the context of motion through a resistive medium. It represents the maximum velocity an object can achieve when falling under the influence of gravity and resistance. Another term often used interchangeably is 'terminal velocity'.

From the steps in the exercise, we determined that the limiting velocity \[v_{lim} = -\frac{mg}{c} \]. This result indicates:
  • When an object is falling, the downward force of gravity \ (mg) \ is balanced by the upward resistive force \( \frac{cv}{m} \).
  • At this point, the object stops accelerating and continues to fall at a constant velocity.
Understanding this helps in various physical scenarios, such as predicting how fast a parachutist might fall before deploying their parachute.

So to wrap it up, the key takeaway is that the limiting velocity is the steady speed an object reaches due to a balance of all acting forces, illustrating the concept of equilibrium in motion.

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Most popular questions from this chapter

(Escape Velocity) Suppose that a rocket is launched from the Earth's surface. At a great (radial) distance \(r\) from the center of the Earth, the rocket's acceleration is not the constant \(g\). Instead, according to Newton's law of gravitation, \(a=k / r^{2}\), where \(k\) is the constant of proportionality ( \(k>0\) if the rocket is falling toward the Earth; \(k<0\) if the rocket is moving away from the Earth). (a) If \(a=-g\) at the Earth's surface (when \(r=R\) ), find \(k\) and show that the rocket's velocity is found by solving the initial value problem \(d v / d t=-g R^{2} / r^{2}, v(0)=v_{0}\). (b) Show that \(d v / d t=v d v / d r\) so that the solution to the initial value problem \(v d v / d r=\) \(-g R^{2} / r^{2}, v(R)=v_{0}\) is \(v^{2}=2 g R^{2} / r+v_{0}^{2}-\) \(2 g R\). (c) If \(v>0\) (so that the rocket does not fall to the ground), show that the minimum value of \(v_{0}\) for which this is true (even for very large values of \(r\) ) is \(v_{0}=\sqrt{2 g R}\). This value is called the escape velocity and signifies the minimum velocity required so that the rocket does not return to the Earth.

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The half-life of \({ }^{14} \mathrm{C}\) is 5730 years. If the original amount of \({ }^{14} \mathrm{C}\) in a particular living organism is \(20 \mathrm{~g}\) and that found in a fossil of that organism is \(0.01 \mathrm{~g}\), determine the approximate age of the fossil.

A projectile of mass \(100 \mathrm{~kg}\) is launched vertically from ground level with an initial velocity of \(100 \mathrm{~m} / \mathrm{s}\). (a) If the projectile is subjected to air resistance equivalent to \(1 / 10\) times the instantaneous velocity, determine the velocity of and the height of the projectile at any time \(t\). (b) What is the maximum height attained by the projectile?

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