Acceleration (AI HL)
Acceleration ties together the three kinematics functions: it's the rate of change of velocity, and the second rate of change of displacement. Most questions here ask you to move between the three, either by differentiating down from \(s\) or \(v\), or by integrating back up from \(a\) once you know a starting velocity. It's part of the broader Kinematics topic.
27 questions on this sub-topic.
The key formulas
Covered under IB syllabus reference AHL5.13: kinematic problems involving displacement \(s\), velocity \(v\) and acceleration \(a\), including \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}=v\dfrac{dv}{ds}\). Both forms below are in the formula booklet.
Acceleration from velocity or displacement
\(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}=v\dfrac{dv}{ds}\)
Differentiate \(v(t)\) once, or \(s(t)\) twice, to get \(a(t)\).
Alternative acceleration form
\[a=v\dfrac{dv}{ds}\]
Useful when acceleration is more naturally expressed as a function of displacement rather than time.
✓ In the formula bookletNeed the full syllabus wording and formula-booklet reference table? See Kinematics.
Worked examples
A particle has \(v(t)=5\sin t\) (m/s).
(a) Find \(a(t)\).
(b) Find the acceleration at \(t=0.\)
Worked solution
(a) \(a(t) = 5\cos t.\) M1
\(a(t) = 5\cos t.\) A1
(b) \(a(0) = 5\) m/s². M1
\(a(0) = 5\) m/s². A1
A particle has acceleration \(a(t) = 4\) m/s\(^2\) (constant). Initially \(v = -6\) m/s and \(s = 2\) m.
(a) Find \(v(t).\)
(b) Find \(s(t).\)
Worked solution
(a) \(v = 4t + C;\ v(0) = -6 \Rightarrow v(t)\) M1
\(= 4t - 6.\) A1
(b) \(s = 2t^2 - 6t + C;\ s(0) = 2 \Rightarrow s(t)\) M1
\(= 2t^2 - 6t + 2.\) A1
A particle has acceleration \(a(t) = 6t - 4\) m/s² and initial velocity \(v(0) = 5\) m/s.
(a) Find \(v(t).\)
(b) Find \(v(3).\)
Worked solution
(a) \(v = \int(6t - 4)\,dt\) M1
\(= 3t^2 - 4t + C.\) A1
\(v(0) = 5 \Rightarrow C = 5.\) A1
(b) \(v(3) = 27 - 12 + 5\) M1
\(= 20\) m/s. A1
Common mistakes
- Going the wrong direction between \(s\), \(v\) and \(a\). Acceleration comes from differentiating, not integrating, velocity - it's easy to panic under time pressure and integrate instead when the question mentions "acceleration".
- Dropping the constant of integration. When you integrate \(a(t)\) to recover \(v(t)\), you must add \(+C\) and use a given condition (often \(v(0)\)) to pin it down - forgetting \(C\) gives a family of answers, not the specific one asked for.
- Losing marks on limiting behaviour. When a question asks for the value \(v\) approaches as \(t\to\infty\), remember that any term like \(e^{-t}\) tends to \(0\), not that the whole expression becomes \(0\) - only substitute the limit of that one term.
Ready to practise properly?
27 acceleration questions, marked instantly like the real exam.
Quick answers
How do you find acceleration from a velocity function?
Differentiate: \(a=\dfrac{dv}{dt}\). If you only have displacement \(s\), differentiate twice, since \(a=\dfrac{d^2s}{dt^2}\).
How do you find velocity from an acceleration function?
Integrate \(a(t)\) with respect to \(t\), then use a known velocity value (often \(v(0)\)) to find the constant of integration.