This page discusses linear motion from the perspective of differential equations and requires a more advanced understanding of calculus including integration. For a beginning look at linear motion and practice problems, see the Linear Motion page in the integrals applications section.

We start this discussion with the differential equation \(\ddot{s} = a\) where \(a\) is acceleration and the independent variable is \(t\) to represent time. The acceleration may be a function of \(t\) or it may be constant. We will assume it is constant. To get the velocity, we integrate this function with respect to \(t\) to get

\( v(t) = \dot{s} = \int{\ddot{s}~dt} = \int{a~dt} = at + c_1 \)

Notice that if we are given an initial condition \(v(0) = v_0\), we have \( v(t) = at + v_0\) where \(v_0\) is the initial velocity.

To get the position function, we integrate the velocity function.

\( s(t) = \int{v(t)~dt} = \int{at+v_0~dt} = \) \( \frac{1}{2}at^2 + v_0 t + c_2 \)

Like we did with velocity, if we are given initial position, \( s(0) = s_0 \), this allows us to solve for \(c_2 = s_0\). We end up with \(s(t) = at^2/2 + v_0 t + s_0\). Now we have all three equations we need, listed in the table below.

position | \(s(t) = at^2/2 + v_0 t + s_0\) |

velocity | \( \dot{s}(t) = at + v_0\) |

acceleration | \( \ddot{s}(t) = a \) |

Okay, that's all you need to solve most of the problems you will come across. Variations may include being given an acceleration function that is not constant. Now you know how to calculate the velocity and position. Practice problems can be found on the integration page dedicated to position, velocity and acceleration.

In this section, we discuss the motion of a falling object, in particular, a special problem where we do not neglect air resistance. We work the problem with lots of detail in several pieces. In order to put it all together, it will help you to rewrite all parts adding any detail you need in order to understand everything.

*Note: This problem is NOT REQUIRED to understand projectile motion. We think it is an interesting problem that will help you understand how differential equations are applied and used and is included here for those of you who will go on and use differential equations in future courses.*

*Problem Statement* - We have an object being dropped out of an airplane of mass \(m\) with initial height \(h_0\). We do not neglect air resistance, which is proportional to the square of the velocity. We want to determine

- the differential equation associated with this motion

- the velocity and position functions

- the velocity as \(t \to \infty \)

- the position function near \( t=0 \)

Position Function |
---|

We assume that the positive direction is up. We have gravity working in the downward direction and wind resistance working against motion and therefore in the positive direction. Our constants and variables are listed below including details on units.

Constants | ||
---|---|---|

symbol |
description |
units |

\(m\) |
mass of the object |
mass |

\(g\) |
gravity |
distance / time |

\(\mu\) |
air friction coefficient |
distance |

Variables | ||

\(t\) |
time |
time |

\(v\) |
velocity |
distance / time |

\(a = v' = \dot{v}\) |
acceleration |
distance / time |

We start with an equation from physics \( F = ma = -mg + \mu v^2 \).

Since acceleration \( a = dv/dt = \dot{v} \), we have \( m\dot{v} = -mg + m \mu v^2 \to \) \( m\dot{v} - m \mu v^2 = -mg \) as our differential equation.

Because we have \(v^2\), this is a nonlinear differential equation and, therefore, not easy to solve. Due to the length and complexity of the solution, the details of finding the velocity function can be found in the next panel. The solution requires a special substitution technique and foreknowledge of the goal. Studying the solution will give you an appreciation and knowledge of the technique and it displays the beauty of mathematics. So it is worth taking some time to work through it.

*Due to the complexity and length of this solution, we have decided to place it here in a separate panel. As a student, it would usually not be expected of you to come up with this solution. This is deep, tricky and has some techniques that you may not have seen before. So, get out a piece of paper and pencil and work through this slowly and carefully.*

We have an object being dropped out of an airplane of mass \(m\). The wind resistance is proportional to the square of the velocity. We want to determine the differential equation associated with this motion and solve for the velocity and position functions. The basic differential equation \( m\dot{v} - m \mu v^2 = -mg \) is set up in the previous panel. In this section, we show that the solution is

\(\displaystyle{ v(t) = -V_T \tanh(\omega t) = V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] }\) where \( \omega = \sqrt{\mu g} \) and \( V_T = \sqrt{g/\mu} \) |

First, we determine where we are going. We start with \( m\dot{v} = m\mu v^2 - mg\) and we divide through by \( m\), which we can do since \( m \neq 0 \), to get \( \dot{v} = \mu v^2 - g\). We are going to use substitution to get \( g\dot{u} = g u^2 - g ~~~ \to ~~~ \dot{u} = u^2 - 1 \). Notice that, even though this is still nonlinear, it is separable, so we should be able to solve it.

This technique requires us to introduce a variable \(u\) which is a function of \(s\). So we have a function \( u(s) \) and we define \(\displaystyle{ \dot{u} = \frac{du}{ds} }\).

So now we compare the two equations, term-by-term.

\(\begin{array}{ccccl}
\displaystyle{\frac{dv}{dt}} & = & \mu v^2 & - & g \\
\displaystyle{g\frac{du}{ds}} & = & g u^2 & - & g \\
\downarrow & & \downarrow \\
\displaystyle{\frac{dv}{dt} = g\frac{du}{ds}} & & \mu v^2 = g u^2 & & ~~~~~ [ 1 ]
\end{array}\)

If we can find a function \( u(s) \), where \(s\) is a function of \(t\), that satisfies these last two equations, we can solve the original differential equations.

Let's start with \(\displaystyle{ \mu v^2 = g u^2 ~~~ \to ~~~ v^2 = \frac{g}{\mu}u^2 ~~~ \to ~~~ v = \sqrt{g/\mu} ~u }\).

We will call the constant \(\displaystyle{ V_T = \sqrt{g/\mu} }\) [ terminal velocity - you will see why later ].

So \( v(t) = V_T u(s) \) where \( V_T \) is a constant. I wrote the functions with their variables to emphasize each variable.

Let's take the derivative of the last equation with respect to \(t\).

\(\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{dt}}\)

Since \( u \) is a function of \(s\), we use the chain rule to get \(\displaystyle{ \frac{du}{dt} = \frac{du}{ds} \frac{ds}{dt} }\).

So now we have \(\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{ds} \frac{ds}{dt} ~~~ [ 2 ] }\)

Now let's go back and use equation [1] \(\displaystyle{ \frac{dv}{dt} = g \frac{du}{ds} }\) and combine it with equation [2] to get

\(\displaystyle{g \frac{du}{ds} = V_T \frac{du}{ds} \frac{ds}{dt} }\)

\(\displaystyle{ g = V_T \frac{ds}{dt} }\)

\(\displaystyle{ \frac{ds}{dt} = \frac{g}{V_T}}\)

Since \(\displaystyle{ V_T = \sqrt{g/\mu} ~~~ \to \frac{ds}{dt} = \frac{g}{\sqrt{g/\mu}} = \sqrt{\mu g}}\).

We will call this constant \( \omega = \sqrt{\mu g} \).

Now we have \(\displaystyle{ \frac{ds}{dt} = \omega ~~~ \to ~~~ ds = \omega dt ~~~ \to ~~~ s = \omega t + k }\); \( k = 0 \) due to the initial conditions.

So we have derived \( s = \omega t \) where \( \omega = \sqrt{\mu g} \).

So, finally, we have derived the equations we need to convert \( \dot{v} = \mu v^2 - g \) to \( \dot{u} = u^2 - 1 \). Let's work forward to check our work.

We will start with \(\displaystyle{ \frac{dv}{dt} = \mu v^2 - g }\) and let \( v(t) = V_T u(s) \) where
\(\displaystyle{ V_T = \sqrt{g/\mu} }\).

\(\displaystyle{ v(t) = V_T u(s) }\)

\(\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{dt} }\)

\(\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{ds} \frac{ds}{dt}}\)

\(\displaystyle{ s = \omega t ~~~ \to ~~~ \frac{ds}{dt} = \omega }\)

Now we will combine \(\displaystyle{ \frac{dv}{dt} = \mu v^2 - g }\) with \(\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{ds} \frac{ds}{dt} }\) to get

\(\displaystyle{ V_T \omega \frac{du}{ds} = \mu V_T^2 u^2 - g }\)

Next, we will replace \( V_T = \sqrt{g/\mu} \) and \( \omega = \sqrt{\mu g} \)

\(\displaystyle{ (\sqrt{g/\mu})(\sqrt{\mu g}) \frac{du}{ds} = \mu (\sqrt{g/\mu})^2 u^2 - g }\)

\(\displaystyle{ g \frac{du}{ds} = gu^2 - g }\)

\(\displaystyle{ \frac{du}{ds} = u^2 - 1 }\)

From the initial conditions, we know that \( v(0) = 0 \) since \( s=\omega t \) so \( s=0 \) when \( t=0 \).

Since \( v(t) = V_T u(s) ~~~ \to ~~~ v(0) = V_T u(0) ~~~ \to ~~~ u(0) = 0 \) when \( s=0 \).

And this is what we wanted.

**Recap**

Let's recap what we just did and where we go from here.

What we've done so far is show how we can use the equations \( v(t) = V_T u(s); ~~~ V_T = \sqrt{g/\mu} ~~~ \omega = \sqrt{\mu g} \)
to convert \(\displaystyle{ \frac{dv}{dt} = \mu v^2 - g }\) to \(\displaystyle{ \frac{du}{ds} = u^2 - 1 }\).
The explanation shows where the (strange) constants \( V_T \) and \(\omega\) come from.

Now our goal is to solve \(\displaystyle{ \frac{du}{ds} = u^2 - 1 }\) and use the result to get \( v(t) = V_T u(s) \) remembering that \( s=\omega t \). So, let's separate the differential equation and solve.

\(\displaystyle{ \frac{du}{ds} = u^2 - 1 ~~~ \to ~~~ \frac{du}{u^2-1} = ds ~~~ \to ~~~ \int{\frac{du}{u^2-1}} = \int{ds} = s + c_1 }\)

The integral \(\displaystyle{ \int{\frac{du}{u^2-1}} }\) is solved on the integration by partial fractions page. The result is

\(\displaystyle{ \int{\frac{du}{u^2-1}} = \frac{1}{2}\ln \abs{ \frac{u-1}{u+1} } + c_2 }\)

So we have \(\displaystyle{ \frac{1}{2}\ln \abs{ \frac{u-1}{u+1} } = s + c_3 }\) where \( c_3 = c_1 - c_2 \).

Now we may be tempted here to solve for \(c_3\) but the absolute values could give us the wrong answer. So do one more step.

\(\displaystyle{ \frac{u-1}{u+1} = (e^{s+c_3})^2 = Ae^{2s} }\) where \( A = e^{2c_3} \).

Now we need to use the initial conditions to determine \(A\).

\(\displaystyle{ u(0) = 0 ~~~ \to ~~~ \frac{0-1}{0+1} = Ae^0 ~~~ \to ~~~ A = -1 }\)

So now we have \(\displaystyle{ \frac{u-1}{u+1} = -e^{2s} }\) and we need to solve for \(u \).

\(\begin{array}{rcl}
u - 1 & = & -e^{2s}(u+1) \\
& = & -e^{2s}u-e^{2s} \\
u + ue^{2s} & = & 1 - e^{2s} \\
u(1+e^{2s}) & = & \\
u & = & \displaystyle{\frac{1-e^{2s}}{1+e^{2s}}}
\end{array}\)

Note: If you are familiar with hyperbolic trig functions, you may have noticed that

\(\displaystyle{ \int{\frac{du}{u^2-1}} = -\tanh^{-1}(u) }\) and so \(\displaystyle{ u = -\tanh(s) = \frac{1-e^{2s}}{1+e^{2s}} }\).

Since \( s = \omega t \), \(\displaystyle{ u(s) = u(\omega t) = -\tanh(\omega t) = \frac{1-e^{2\omega t}}{1+e^{2\omega t}} }\).

Substituting this result into \( v(t) = V_T u(s) \), we finally obtain our result

\(\displaystyle{ v(t) = -V_T \tanh(\omega t) = V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] }\) where \( \omega = \sqrt{\mu g} \) and \( V_T = \sqrt{g/\mu} \) |

From the solution in the panel, we have a couple of special constants, \( \omega = \sqrt{\mu g} \) and \( V_T = \sqrt{g/\mu} \). So we start with the result of that panel where the velocity function is

\(\displaystyle{v(t) = -V_T ~ \tanh(\omega t) = V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right]}\)

Velocity As \( t \to \infty \) |
---|

Let's look at what happens to the velocity as \(t\) approaches infinity by taking the limit

\(\displaystyle{
\lim_{t \to \infty}{V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] } = }\) \(\displaystyle{
\lim_{t \to \infty}{V_T \left[ \frac{-2\omega e^{2\omega t}}{2\omega e^{2\omega t}} \right] } =
-V_T
}\)

[We used L'Hôpital's Rule here.]

So as \( t \to \infty\) the velocity \( v(t) \to -V_T = -\sqrt{g/\mu} \), which is called the terminal velocity.

Position Function Near \( t=0 \) |
---|

We need to integrate the velocity function to obtain the position function, i.e. we need to solve
\(\displaystyle{ \frac{dy}{dt} = v(t) }\) and \( y(0) = h_0 \) is the initial height.

\(\displaystyle{ y(t) = \int{v(t)~dt} = \int{-V_t \tanh(\omega t)~dt} }\)

The details for solving the integral \(\displaystyle{ \int{\tanh(\omega t)~dt} }\) can be found as a practice problem on the integrals page.

So we have \(\displaystyle{ y(t) = \frac{-V_T}{\omega}\ln\abs{\cosh(\omega t)} + k_1 }\). Using the initial condition, we end up with
\(\displaystyle{ y(t) = \frac{-V_T}{\omega}\ln\abs{\cosh(\omega t)} + h_0 }\)

In order to determine what is going on very near \( t=0\) we replace \(\displaystyle{ \frac{V_T}{\omega}\ln\abs{\cosh(\omega t)} }\) with it's second degree Taylor series polynomial. The details can be found as a practice problem on the Taylor Series page with the result

\(\displaystyle{ \frac{V_T}{\omega}\ln\abs{\cosh(\omega t)} \approx \frac{\omega^2 t^2}{2\mu} }\)

This gives us \(\displaystyle{ y(t) \approx h_0 - \frac{\omega^2 t^2}{2\mu} = }\) \(\displaystyle{
h_0 - \frac{\mu g t^2}{2\mu} = }\) \(\displaystyle{
h_0 - \frac{1}{2}gt^2}\)
for small *t*.

Summary of Final Answers | |
---|---|

differential equation |
\(m\dot{v} - m \mu v^2 = -mg\) |

velocity function |
\(\displaystyle{ v(t) = -V_T ~ \tanh(\omega t) = }\) \(\displaystyle{ V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] }\) |

velocity as \(t \to \infty \) |
\(\displaystyle{ \lim_{t \to \infty}{v(t)} = -V_T }\) |

position function |
\(\displaystyle{ y(t) = \frac{-V_T}{\omega}\ln\abs{\cosh(\omega t)} + h_0 }\) |

position function near \( t=0 \) |
\(\displaystyle{ y(t) \approx h_0 - \frac{1}{2}gt^2}\) |

In the process of this solution, we defined the following intermediate values.

\( \omega = \sqrt{\mu g} \) |
\( V_T = \sqrt{g/\mu} \) |

Work this practice problem based on this discussion.

You CAN Ace Differential Equations

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