Variable forces
Study variable forces for Year 12 Specialist Mathematics in Queensland (QCAA). When the resultant force on a particle changes with time, position or velocity, you apply Newton's second law and integrate to describe the motion.
You will learn to choose the right form of the acceleration, form and solve a differential equation of motion, and find velocity as a function of time or position, the distance a resisted body travels, and its terminal velocity — the core of modelling motion in a straight line.
Theory
Variable forces extend Newton's second law \(F=ma\) to a resultant force that changes with time, position or velocity, in Year 12 Specialist Mathematics (QCAA, Queensland). The skill is choosing the right form of the acceleration — \(\dfrac{dv}{dt}\), \(v\dfrac{dv}{dx}\) or \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) — then forming a differential equation and integrating to find \(v\).
When the resultant force on a particle is not constant, its acceleration is not constant either, so the constant-acceleration formulas no longer apply. Instead you apply Newton's second law \(F=ma\) with \(F\) written as a function of \(t\), \(x\) or \(v\), and solve the resulting differential equation.
The acceleration has three equivalent forms: \(a=\dfrac{dv}{dt}\) (rate of change of velocity with time), \(a=v\dfrac{dv}{dx}\) (from the chain rule), and \(a=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\). Choosing the form whose variables match the force is what makes the integral separable.
Use \(a=\dfrac{dv}{dt}\) when the force depends on time; use \(a=v\dfrac{dv}{dx}\) when the force depends on position; and for a force depending on velocity (such as air resistance) use \(\dfrac{dv}{dt}\) to reach \(v(t)\) or \(v\dfrac{dv}{dx}\) to reach \(v(x)\).
Two special outcomes appear often: a terminal velocity, the constant speed where the acceleration falls to zero, and a limiting speed reached as \(x\to\infty\). This sub-topic excludes pulleys and connected bodies.
Newton's second law with a variable resultant force, and the three forms of the acceleration for straight-line motion:
Separating variables gives the working integrals for each case:
The terminal velocity of a falling body is found by setting the acceleration to zero:
Solving a variable-force problem
- Apply \(F=ma\) to get \(a=\dfrac{F}{m}\), and note whether the force depends on \(t\), \(x\) or \(v\).
- Choose the acceleration form: \(\dfrac{dv}{dt}\) for a force in \(t\) (or for \(v(t)\)); \(v\dfrac{dv}{dx}\) for a force in \(x\) (or for \(v(x)\)).
- Separate and integrate both sides to form the general solution, keeping the constant of integration.
- Apply the initial condition to find the constant, then substitute the required value (or take a limit for a terminal or limiting speed).
The force depends on \(t\), so use \(a=\dfrac{dv}{dt}=\dfrac{F}{m}\):
| \(a\) | \(=\) | \(\dfrac{12t}{2}\) |
| \(=\) | \(6t\) |
Integrate \(\dfrac{dv}{dt}=6t\) with \(v(0)=0\):
| \(v\) | \(=\) | \(\int 6t\,dt\) |
| \(=\) | \(3t^2+c\) | |
| \(v(0)=0\) | \(\Rightarrow\) | \(c=0\) |
| \(v\) | \(=\) | \(3t^2\) |
Integrate \(v=\dfrac{dx}{dt}\) with \(x(0)=0\), then substitute \(t=3\):
| \(x\) | \(=\) | \(\int 3t^2\,dt\) |
| \(=\) | \(t^3\) | |
| \(x(3)\) | \(=\) | \(3^3\) |
| \(=\) | \(27\) |
\(v=3t^2\text{ m/s}\) and the displacement after \(3\text{ s}\) is \(27\text{ m}\).
The force depends on \(x\), so use \(a=v\dfrac{dv}{dx}=\dfrac{F}{m}=8x\):
| \(v\dfrac{dv}{dx}\) | \(=\) | \(8x\) |
Separate and integrate both sides:
| \(\int v\,dv\) | \(=\) | \(\int 8x\,dx\) |
| \(\dfrac{v^2}{2}\) | \(=\) | \(4x^2+c\) |
| \(v^2\) | \(=\) | \(8x^2+c_1\) |
Apply \(v=0\) at \(x=1\) to find \(c_1\), then substitute \(x=3\):
| \(0\) | \(=\) | \(8(1)^2+c_1\) |
| \(c_1\) | \(=\) | \(-8\) |
| \(v^2\) | \(=\) | \(8x^2-8\) |
| \(v^2(3)\) | \(=\) | \(8(9)-8\) |
| \(=\) | \(64\) | |
| \(v\) | \(=\) | \(8\) |
\(v=8\text{ m/s}\) as the particle passes \(x=3\text{ m}\).
The force depends on \(v\); for \(v(t)\) use \(a=\dfrac{dv}{dt}=8-2v\) and separate:
| \(\int\dfrac{dv}{8-2v}\) | \(=\) | \(\int 1\,dt\) |
| \(-\tfrac12\ln(8-2v)\) | \(=\) | \(t+c\) |
Rearrange and apply \(v(0)=0\):
| \(8-2v\) | \(=\) | \(Ae^{-2t}\) |
| \(v(0)=0\) | \(\Rightarrow\) | \(A=8\) |
| \(8-2v\) | \(=\) | \(8e^{-2t}\) |
| \(v\) | \(=\) | \(4\left(1-e^{-2t}\right)\) |
Terminal velocity: set \(a=0\) (equivalently let \(t\to\infty\)):
| \(8-2v\) | \(=\) | \(0\) |
| \(v\) | \(=\) | \(4\) |
\(v=4\left(1-e^{-2t}\right)\text{ m/s}\), rising to a terminal velocity of \(4\text{ m/s}\).
For a distance, use \(a=v\dfrac{dv}{dx}=-\dfrac{6v}{2}=-3v\):
| \(v\dfrac{dv}{dx}\) | \(=\) | \(-3v\) |
| \(\dfrac{dv}{dx}\) | \(=\) | \(-3\) |
Integrate and apply \(v=15\) at \(x=0\):
| \(v\) | \(=\) | \(-3x+c\) |
| \(v(0)=15\) | \(\Rightarrow\) | \(c=15\) |
| \(v\) | \(=\) | \(15-3x\) |
The body stops when \(v=0\):
| \(0\) | \(=\) | \(15-3x\) |
| \(3x\) | \(=\) | \(15\) |
| \(x\) | \(=\) | \(5\) |
The body travels \(5\text{ m}\) before coming to rest.
Common pitfalls
Frequently asked questions
When do you use v dv/dx instead of dv/dt?
Use \(a=v\dfrac{dv}{dx}\) when the force depends on position \(x\), or when you want the velocity as a function of position. Use \(a=\dfrac{dv}{dt}\) when the force depends on time, or when you want velocity as a function of time.
Why can't I use v = u + at for a variable force?
The constant-acceleration formulas are derived assuming \(a\) is constant. A variable force gives a variable acceleration, so instead you write \(a=\dfrac{dv}{dt}\) or \(v\dfrac{dv}{dx}\), form a differential equation, and integrate.
How do you find a terminal velocity?
Terminal velocity is the steady speed a falling body approaches, where the resultant force and hence the acceleration are zero. Set \(a=0\) in the equation of motion and solve for \(v\).
What is the point of the form d/dx of half v squared?
It is just \(v\dfrac{dv}{dx}\) written another way, since \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)=v\dfrac{dv}{dx}\). It makes integrating with respect to \(x\) quick, because \(\tfrac12 v^2\) is the antiderivative directly.
How do you get the distance a resisted body travels before stopping?
Use \(a=v\dfrac{dv}{dx}\) so the equation is in \(v\) and \(x\), integrate to get \(v\) as a function of \(x\), then set \(v=0\) and solve for \(x\).
Do I always need the initial condition?
Yes. Integrating introduces a constant, and the initial condition — the speed at a known time or position — is what determines it. Without it the velocity function is not fully known.