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Chapter 10: Problem 16

A skydiver's vertical velocity is governed by the differential equation $$ m \frac{\mathrm{d} v}{\mathrm{~d} t}=m g-K v^{2} $$ where \(K\) is the skydiver's coefficient of drag. If the skydiver leaves her aeroplane at time \(t=0\) with zero vertical velocity, find at what time she reaches half her final velocity.

Short Answer

Expert verified
Time to reach half the final velocity is \( \frac{m}{\sqrt{K} g} \tan^{-1}\left( \frac{1}{2\sqrt{K/mg}} \right) \)."

Step by step solution

01

Identify the Differential Equation

The given differential equation is \( m \frac{\mathrm{d} v}{\mathrm{~d} t} = m g - K v^2 \). This is a separable differential equation representing the motion of the skydiver.
02

Find the Final Velocity

To find the final velocity, set \( \frac{\mathrm{d} v}{\mathrm{~d} t} = 0 \) since the velocity becomes constant at terminal velocity. Solving \( m g - K v_T^2 = 0 \) gives the terminal velocity \( v_T = \sqrt{\frac{m g}{K}} \).
03

Solve the Differential Equation

Rearrange the equation to \( \frac{\mathrm{d} v}{m g - K v^2} = \frac{\mathrm{d} t}{m} \) and integrate both sides. The integral \( \int \frac{1}{m g - K v^2} \, \mathrm{d} v \) is solved using partial fraction decomposition or a trigonometric substitution.
04

Integration and Substitution

Using substitution \( v = v_T \sin(\theta) \), integrate the left side to \( \frac{1}{\sqrt{K}} \tan^{-1}\left( \frac{v}{v_T \sqrt{K/mg}} \right) \), equivalent with our limits of zero initial velocity and half terminal velocity.
05

Solve for Time

Solving the integrated expression for \( t \) when \( v = \frac{1}{2}v_T \), involves finding the corresponding angle and time: \( t = \frac{m}{\sqrt{K} g} \tan^{-1}\left( \frac{1}{2\sqrt{K/mg}} \right) \).

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a specific type of differential equation where variables can be separated on either side of the equation. This means that one variable can be isolated with its differential on one side and the other variable can be isolated with its differential on the other side.
For example, in the context of our skydiver's velocity equation:
  • The equation is given as \( m \frac{\mathrm{d} v}{\mathrm{~d} t} = m g - K v^2 \).
  • The goal is to separate variables so we can integrate each side separately.
To achieve this, we rearrange terms by moving all instances of \( v \) to one side and \( t \) to the other:
  • This involves dividing both sides by \( m g - K v^2 \) and multiplying by \( \mathrm{d} t \).
  • We end up with: \( \frac{\mathrm{d} v}{m g - K v^2} = \frac{\mathrm{d} t}{m} \).
Once rearranged, the integration process can begin. This allows us to find a more general solution for \( v(t) \), rather than just a rate of change.
Terminal Velocity
Terminal velocity is a fundamental concept in physics referring to the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration.
In the example of the skydiver, terminal velocity is reached when the gravitational force pulling them downwards is exactly balanced by the drag force acting upwards.
  • The force of gravity is expressed as \( mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity.
  • The drag force is proportional to the square of velocity, \( Kv^2 \), where \( K \) is the drag coefficient.
Setting these two forces equal gives the terminal velocity:
  • \( mg - Kv^2 = 0 \).
  • Solving for \( v_T \), we find \( v_T = \sqrt{\frac{mg}{K}} \).
Understanding terminal velocity is crucial in studying motions where objects meet resistance from a fluid (air, water, etc.). The concept ensures the skydiver no longer accelerates and stabilizes at a predictable, constant speed.
Integration Techniques
Integration techniques are strategies used to solve complex integrals by simplifying them into a solvable form. Specifically, for our exercise:
The integral \( \int \frac{1}{m g - K v^2} \, \mathrm{d} v \) appears, suggesting a need for techniques like substitution or partial fraction decomposition.
  • Substitution: You let \( v = v_T \sin(\theta) \) to transform the integral into a simpler trigonometric form.
  • This change of variables matches the inside of the integral to a form you know how to integrate.
  • Solving the resulting trigonometric integral often involves \( \tan^{-1} \), which correlates back to \( \frac{v}{v_T \sqrt{K/mg}} \).
Integrating these types of functions allows us to relate velocity at various points in time to known values, such as half of the terminal velocity. Using these methods guides us through from the beginning of skydiving motion equations to interpretations of real-world velocities.

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Most popular questions from this chapter

Find the damping parameters and natural frequencies of the systems governed by the following second-order linear constant-coefficient differential equations: (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 a \frac{\mathrm{d} x}{\mathrm{~d} t}+16 p^{2} x=0\) (b) \(2 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+14 \frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{1}{\alpha} x=0\) (c) \(2.41 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+1.02 \frac{\mathrm{d} x}{\mathrm{~d} t}+7.63 x=0\) (d) \(\frac{1}{\eta} \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+40 \frac{\mathrm{d} x}{\mathrm{~d} t}+25 \eta x=0\) (e) \(1.88 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4.71 \frac{\mathrm{d} x}{\mathrm{~d} t}+0.48 x=0\)

Find the solution of the following initial-value problem: $$ x^{3} t \frac{\mathrm{d} x}{\mathrm{~d} t}=t^{4}+x^{4}, \quad x(1)=4 $$

For each of the following sets of linearly dependent functions find \(k_{1}, k_{2}, \ldots\) such that \(k_{1} f_{1}+k_{2} f_{2}+\ldots=0\) (a) \(\\{\sin t, \cos t+\sin t, \cos 2 t-\sin t, \cos t-\cos 2 t\\}\) (b) \(\left\\{t+t^{3}, t-t^{2}, t^{2}+2 t^{3}, t^{2}-t^{3}\right\\}\) (c) \(\left\\{\ln t, \ln 2 t, \ln 4 t^{2}\right\\}\) (d) \(\\{f(t)+g(t), f(t)(1+f(t)), g(t)-f(t)\) \(\left.f(t)^{2}-g(t)\right\\}\) (e) \(\left\\{1+t+2 t^{2}, t-2 t^{2}+3 t^{3}, 1+t-2 t^{2},\right.\), \(\left.t-2 t^{2}-3 t^{3}, t^{3}\right\\}\)

For each of the following differential equations determine whether they are exact equations and, if so, find the general solutions: (a) \(x \frac{\mathrm{d} x}{\mathrm{~d} t}+t=0\) (b) \(x \frac{\mathrm{d} x}{\mathrm{~d} t}-t=0\) (c) \((x+t) \frac{\mathrm{d} x}{\mathrm{~d} t}+x-t=0\) (d) \(\left(x-t^{2}\right) \frac{\mathrm{d} x}{\mathrm{~d} t}-2 x t=0\) (e) \((x-t) \frac{\mathrm{d} x}{\mathrm{~d} t}-x+t-1=0\) (f) \((2 x+t) \frac{\mathrm{d} x}{\mathrm{~d} t}+x+2 t=0\)

Which of the following two sets are linearly dependent and which are linearly independent? (a) \(\left\\{1, t, t^{2}, t^{3}, t^{4}, t^{5}, t^{6}\right\\}\) (b) \(\left\\{1+t, t^{2}, t^{2}-t, 1-t^{2}\right\\}\)
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