Other expressions for acceleration
Master the other expressions for acceleration in Year 12 VCE Specialist Mathematics. For a particle moving in a straight line the acceleration can be written three equivalent ways — \(\dfrac{dv}{dt}\), \(v\dfrac{dv}{dx}\) and \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\) — all describing the same rate of change of velocity. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Kinematics: rectilinear motion topic of Unit 4.
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 \(\dfrac{dv}{dt}\) when acceleration depends on time and the displacement forms when it depends on position — a key modelling-motion skill.
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 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.