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Year 12 Maths Advanced (2027) Applications of calculus

Velocity & Acceleration as Derivatives

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Applications of calculus

Velocity and acceleration as derivatives extend motion analysis: acceleration is the derivative of velocity, \(a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}\).

Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Applications of calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

Acceleration is the derivative of velocity and the second derivative of displacement. This Year 12 Mathematics Advanced topic (MAV-12-06) links the three.

\(v=\dfrac{dx}{dt}\) and \(a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}\) (also written \(\ddot{x}\)). Differentiate displacement twice, or velocity once, to get acceleration.

Motion in a lineA displacement curve; its first and second derivatives give velocity and acceleration. xy x(t)
Differentiate \(x(t)\) once for \(v\), twice for \(a\).
\[v=\dfrac{dx}{dt},\qquad a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}\]
velocity is dx dt; acceleration is dv dt equals d squared x dt squared

Method

  1. Differentiate \(x\) once for \(v\).
  2. Differentiate again for \(a\).
  3. Substitute the time if a value is asked.
Example 1 — From displacement
\(x=t^{3}\). Find the acceleration at \(t=2\).
Solution
\(v\)\(=\)\(3t^{2}\)
\(a\)\(=\)\(6t\Rightarrow a(2)=12\)
Example 2 — Constant
\(x=t^{2}+3t\). Find the acceleration.
Solution
\(v\)\(=\)\(2t+3\)
\(a\)\(=\)\(2\)
Example 3 — From velocity
\(v=4t^{2}\). Find the acceleration at \(t=1\).
Solution
\(a\)\(=\)\(8t\Rightarrow a(1)=8\)
Example 4 — Full chain
\(x=t^{3}-3t^{2}+2t\). Find \(v\) and \(a\).
Solution
\(v\)\(=\)\(3t^{2}-6t+2\)
\(a\)\(=\)\(6t-6\)

Common pitfalls

Acceleration is a second derivative of displacement.
From velocity, differentiate once.
Constant velocity has zero acceleration.

Frequently asked questions

What is acceleration in terms of derivatives?

It is the derivative of velocity, and the second derivative of displacement, written d squared x by dt squared or x double dot.

How do you find acceleration from displacement?

Differentiate the displacement twice.

What is the notation for acceleration?

It is written d squared x by dt squared, or with two dots over x.

If velocity is constant, what is the acceleration?

Zero, because the velocity is not changing.