Differential equations for motion
Master differential equations for motion for Year 12 Specialist Mathematics in Queensland (QCAA). When a particle moves in a straight line with non-constant acceleration given as a function of time, displacement or velocity, its motion is a first-order differential equation solved by separation of variables.
You will learn to choose the right form of the acceleration \(-\) \(\dfrac{dv}{dt}\), \(v\dfrac{dv}{dx}\) or \(\dfrac{d}{dx}(\tfrac12 v^2)\) \(-\) separate the variables and integrate to find velocity, displacement, time and terminal velocity, the key modelling tool for resisted motion and vertical motion under gravity.
Theory
Differential equations for motion describe straight-line motion in Year 12 Specialist Mathematics (QCAA, Queensland) when the acceleration is a function of time \(t\), displacement \(x\) or velocity \(v\). Because \(a=\dfrac{dv}{dt}=v\dfrac{dv}{dx}=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\), each case becomes a first-order equation solved by separation of variables. This page shows how to choose the right form and integrate to find velocity, displacement, time and terminal velocity.
When a particle moves in a straight line with non-constant acceleration, the acceleration \(a\) is often given as a function of the time \(t\), the displacement \(x\), or the velocity \(v\). Since \(a=\dfrac{dv}{dt}\) is itself a derivative, each case is a first-order differential equation that separation of variables can solve.
Acceleration has three equivalent forms: \(a=\dfrac{dv}{dt}=v\dfrac{dv}{dx}=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\). Choosing the form that matches the variables in the problem is the whole skill \(-\) a poor choice leaves an integral you cannot evaluate.
If \(a\) is a function of \(x\), use \(a=v\dfrac{dv}{dx}\); the equation separates to \(\displaystyle\int v\,dv=\int f(x)\,dx\), giving \(\tfrac12 v^2\) and hence \(v\) as a function of \(x\). If \(a\) is a function of \(t\), use \(a=\dfrac{dv}{dt}\) and integrate with respect to \(t\).
If \(a\) is a function of \(v\), pick the form by what you want to find: \(\dfrac{dv}{dt}=f(v)\) separates to give the time \(t\), while \(v\dfrac{dv}{dx}=f(v)\) separates to give the distance \(x\). The terminal velocity is the constant speed reached when \(a=0\).
The acceleration of a particle in a straight line can be written in three equivalent ways:
When the acceleration is a function of \(x\), use \(v\dfrac{dv}{dx}\) and integrate:
When the acceleration is a function of \(v\), separate for the time or the distance:
How to solve a differential equation for motion
- Identify the variable the acceleration depends on: is \(a\) a function of \(t\), of \(x\), or of \(v\)?
- Choose the matching form: \(\dfrac{dv}{dt}\) for \(a=f(t)\); \(v\dfrac{dv}{dx}\) for \(a=f(x)\); for \(a=f(v)\) use \(\dfrac{dv}{dt}\) to find time or \(v\dfrac{dv}{dx}\) to find distance.
- Separate and integrate both sides, adding a single constant of integration.
- Apply the conditions (initial velocity, starting point, or \(a=0\) for terminal velocity) to evaluate the constant and answer the question.
\(a\) depends on \(x\), so use \(a=v\dfrac{dv}{dx}\) and integrate:
| \(v\dfrac{dv}{dx}\) | \(=\) | \(9x\) |
| \(\int v\,dv\) | \(=\) | \(\int 9x\,dx\) |
| \(\dfrac{1}{2}v^2\) | \(=\) | \(\dfrac{9}{2}x^2+C\) |
Apply \(v=0\) at \(x=0\) to fix \(C\), then take the square root:
| \(v=0\text{ at }x=0\) | \(\Rightarrow\) | \(C=0\) |
| \(v^2\) | \(=\) | \(9x^2\) |
| \(v\) | \(=\) | \(3x\) |
The velocity is \(v=3x\).
\(a\) depends on \(x\); integrate \(v\dfrac{dv}{dx}=6x+4\):
| \(\int v\,dv\) | \(=\) | \(\int (6x+4)\,dx\) |
| \(\dfrac{1}{2}v^2\) | \(=\) | \(3x^2+4x+C\) |
Use \(v=3\) at \(x=0\) so \(\tfrac12(3)^2=C\), then substitute \(x=2\):
| \(C\) | \(=\) | \(\dfrac{9}{2}\) |
| \(v^2\) | \(=\) | \(6x^2+8x+9\) |
| \(\text{at }x=2:\ v^2\) | \(=\) | \(6(4)+8(2)+9\) |
| \(=\) | \(49\) | |
| \(v\) | \(=\) | \(7\) |
The speed is \(v=7\) m/s.
Distance is wanted, so use \(v\dfrac{dv}{dx}\) and divide by \(v\):
| \(v\dfrac{dv}{dx}\) | \(=\) | \(-6v\) |
| \(\dfrac{dv}{dx}\) | \(=\) | \(-6\) |
| \(v\) | \(=\) | \(42-6x\) |
The puck stops when \(v=0\):
| \(0\) | \(=\) | \(42-6x\) |
| \(6x\) | \(=\) | \(42\) |
| \(x\) | \(=\) | \(7\) |
The puck travels \(7\) m before stopping.
Time is wanted, so separate \(\dfrac{dv}{dt}=-\dfrac{1}{3}v\):
| \(\int \dfrac{1}{v}\,dv\) | \(=\) | \(\int -\dfrac{1}{3}\,dt\) |
| \(\ln v\) | \(=\) | \(-\dfrac{1}{3}t+C\) |
| \(v\) | \(=\) | \(Ae^{-t/3}\) |
Apply \(v=15\) at \(t=0\), then substitute \(t=6\):
| \(t=0,\ v=15\) | \(\Rightarrow\) | \(A=15\) |
| \(v\) | \(=\) | \(15e^{-t/3}\) |
| \(\text{at }t=6:\ v\) | \(=\) | \(15e^{-2}\) |
| \(\approx\) | \(2.03\) |
\(v=15e^{-t/3}\); at \(t=6\) s, \(v\approx2.03\) m/s.
Common pitfalls
Frequently asked questions
What is a differential equation for motion?
It is an equation linking the acceleration of a particle to its velocity, displacement or time. Because \(a=\dfrac{dv}{dt}=v\dfrac{dv}{dx}\), writing \(a\) as a function of \(t\), \(x\) or \(v\) gives a first-order differential equation you solve by separating the variables and integrating.
Which form of acceleration should I use?
Match the form to the variable \(a\) depends on. Use \(\dfrac{dv}{dt}\) when \(a=f(t)\); use \(v\dfrac{dv}{dx}\) when \(a=f(x)\). When \(a=f(v)\), use \(\dfrac{dv}{dt}\) to find the time and \(v\dfrac{dv}{dx}\) to find the distance.
What does \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) mean?
It is another way of writing \(v\dfrac{dv}{dx}\), because differentiating \(\tfrac12 v^2\) with respect to \(x\) by the chain rule gives \(v\dfrac{dv}{dx}\). It is handy when the acceleration depends on \(x\), since integrating it returns \(\tfrac12 v^2\) directly.
How do I find terminal velocity?
Set the acceleration to zero. For a resisted fall such as \(a=g-kv\) or \(a=g-kv^2\), solving \(a=0\) for \(v\) gives the terminal (limiting) velocity \(-\) the constant speed the object approaches. No integration is required.
Why do I sometimes get \(v\) as a function of \(x\) and sometimes of \(t\)?
It depends on the form you integrate. Using \(v\dfrac{dv}{dx}\) integrates with respect to \(x\), so you get \(v\) in terms of \(x\). Using \(\dfrac{dv}{dt}\) integrates with respect to \(t\), so you get \(v\) in terms of \(t\).
When can I use these differential-equation methods?
Use them whenever the acceleration is non-constant \(-\) a function of \(t\), \(x\) or \(v\). If the acceleration is constant, the simpler constant-acceleration formulas apply instead.