Other expressions for acceleration
Master the other expressions for acceleration in Year 12 Specialist Mathematics for Queensland (QCAA). For a particle moving in a straight line the acceleration can be written three equivalent ways, dv/dt, v dv/dx and d/dx of one half v squared, all describing the same rate of change of velocity.
You will see where the displacement forms come from through the chain rule, and learn to choose the right form for the information given — using dv/dt when acceleration depends on time and the displacement forms when it depends on position — a key modelling-motion skill in Unit 4.
Theory
For a particle moving in a straight line, the acceleration can be written three equivalent ways: \(a=\dfrac{dv}{dt}=v\dfrac{dv}{dx}=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\). This page of Year 12 Specialist Mathematics (QCAA, Queensland) shows where these forms come from and how to choose the right one for the information you are given — \(\dfrac{dv}{dt}\) when the acceleration depends on time, and the displacement forms when it depends on position.
A particle moving in a straight line has displacement \(x\), velocity \(v=\dfrac{dx}{dt}\) and acceleration \(a\). Acceleration is the rate of change of velocity, so the most familiar form is \(a=\dfrac{dv}{dt}\) — the derivative of velocity with respect to time.
Often, though, the acceleration or the velocity is given as a function of the displacement \(x\) rather than of time. Then \(\dfrac{dv}{dt}\) is awkward, because \(v\) is not written in terms of \(t\). Two further expressions for the same acceleration solve this: \(a=v\dfrac{dv}{dx}\) and \(a=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\).
All three are the same acceleration. The chain rule links them: since \(v\) depends on \(x\) and \(x\) depends on \(t\), \(a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=v\dfrac{dv}{dx}\). Differentiating \(\tfrac12 v^2\) then gives \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)=v\dfrac{dv}{dx}\), so this last form is just a tidy way of writing \(v\dfrac{dv}{dx}\) that is easy to integrate.
The choice of form is driven by the information. Use \(a=\dfrac{dv}{dt}\) when the acceleration is a function of time; use \(a=v\dfrac{dv}{dx}\) or \(a=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) when the acceleration or velocity is a function of displacement. Picking the matching form is what makes the problem solvable in one clean step.
The acceleration of a particle moving in a straight line has three equivalent forms:
The displacement forms come from the chain rule, because \(v\) is a function of \(x\) and \(x\) is a function of \(t\):
Differentiating \(\tfrac12 v^2\) with respect to \(x\) recovers the same thing, which is why the third form is so useful for integrating:
Choosing and using the right form
- Identify the variable: read whether the acceleration (or velocity) is given as a function of time \(t\), displacement \(x\), or velocity \(v\).
- Choose the form: use \(a=\dfrac{dv}{dt}\) when \(a=f(t)\); use \(a=v\dfrac{dv}{dx}\) or \(a=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) when \(a\) or \(v\) is a function of \(x\).
- Apply it: for \(a=v\dfrac{dv}{dx}\), differentiate \(v(x)\) and multiply by \(v\); to find \(v^2\) from \(a=f(x)\), write \(a=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) and integrate once.
- Fix the constant from the initial condition (a known velocity at a known displacement), then answer what is asked.
The velocity is a function of \(x\), so use \(a=v\dfrac{dv}{dx}\); differentiate first:
| \(\dfrac{dv}{dx}\) | \(=\) | \(5\) |
| \(a\) | \(=\) | \(v\dfrac{dv}{dx}\) |
| \(=\) | \((5x-2)(5)\) | |
| \(=\) | \(25x-10\) |
Substitute \(x=3\):
| \(a\big|_{x=3}\) | \(=\) | \(25(3)-10\) |
| \(=\) | \(65\) |
\(a=25x-10\ \text{m/s}^2\); when \(x=3\), \(a=65\ \text{m/s}^2\).
The acceleration is the derivative of \(\tfrac12 v^2\) with respect to \(x\):
| \(a\) | \(=\) | \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) |
| \(=\) | \(\dfrac{d}{dx}(x^3+2x)\) | |
| \(=\) | \(3x^2+2\) |
Substitute \(x=2\):
| \(a\big|_{x=2}\) | \(=\) | \(3(2)^2+2\) |
| \(=\) | \(12+2\) | |
| \(=\) | \(14\) |
\(a=3x^2+2\ \text{m/s}^2\); when \(x=2\), \(a=14\ \text{m/s}^2\).
Acceleration is a function of \(x\), so use \(a=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) and integrate once:
| \(\dfrac12 v^2\) | \(=\) | \(\textstyle\int (4x-4)\,dx\) |
| \(=\) | \(2x^2-4x+C\) |
Use \(v=0\) at \(x=1\) to fix \(C\), then double:
| \(0\) | \(=\) | \(2(1)^2-4(1)+C\) |
| \(C\) | \(=\) | \(2\) |
| \(\dfrac12 v^2\) | \(=\) | \(2x^2-4x+2\) |
| \(v^2\) | \(=\) | \(4x^2-8x+4\) |
| \(=\) | \(4(x-1)^2\) |
The speed when \(x=4\):
| \(v^2\) | \(=\) | \(4(4-1)^2\) |
| \(=\) | \(36\) | |
| \(v\) | \(=\) | \(6\) |
\(v^2=4(x-1)^2\); the speed when \(x=4\) is \(6\ \text{m/s}\).
Differentiate \(v=(25-x^2)^{1/2}\) by the chain rule:
| \(\dfrac{dv}{dx}\) | \(=\) | \(\tfrac12(25-x^2)^{-1/2}(-2x)\) |
| \(=\) | \(\dfrac{-x}{\sqrt{25-x^2}}\) |
Form \(a=v\dfrac{dv}{dx}\); the surd cancels:
| \(a\) | \(=\) | \(\sqrt{25-x^2}\times\dfrac{-x}{\sqrt{25-x^2}}\) |
| \(=\) | \(-x\) |
Substitute \(x=4\):
| \(a\big|_{x=4}\) | \(=\) | \(-4\) |
\(a=-x\ \text{m/s}^2\); when \(x=4\), \(a=-4\ \text{m/s}^2\).
Common pitfalls
Frequently asked questions
Why are there three expressions for acceleration?
They are three ways of writing the same rate of change of velocity. \(\dfrac{dv}{dt}\) differentiates with respect to time, while \(v\dfrac{dv}{dx}\) and \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) differentiate with respect to displacement — useful when the motion is described in terms of \(x\) instead of \(t\).
When do I use \(v\dfrac{dv}{dx}\) instead of \(\dfrac{dv}{dt}\)?
Use \(v\dfrac{dv}{dx}\) (or \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\)) whenever the acceleration or velocity is given as a function of the displacement \(x\). Use \(\dfrac{dv}{dt}\) when it is given as a function of time \(t\).
Where does \(v\dfrac{dv}{dx}\) come from?
From the chain rule. Since \(v\) depends on \(x\) and \(x\) depends on \(t\), \(\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}\); and \(\dfrac{dx}{dt}=v\), so \(a=v\dfrac{dv}{dx}\).
What is \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) equal to?
It equals \(v\dfrac{dv}{dx}\), which is the acceleration. Differentiating \(\tfrac12 v^2\) gives \(\tfrac12\cdot 2v\cdot\dfrac{dv}{dx}=v\dfrac{dv}{dx}\).
How do I find velocity when acceleration depends on displacement?
Write \(a=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) and integrate both sides with respect to \(x\) to get \(\tfrac12 v^2\). Then use the initial condition to find the constant, and solve for \(v^2\) or \(v\).
Why does \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) make integration easier?
Because it is already a derivative with respect to \(x\). If \(a=f(x)\), then \(\tfrac12 v^2=\int f(x)\,dx\) directly, whereas \(v\dfrac{dv}{dx}=f(x)\) needs you to separate the variables first.