Non-Constant Acceleration
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
The familiar equations (v = v0 + at and the rest) are only true when a is constant. When acceleration changes with time, position or velocity, go back to calculus: v = v0 + ∫a dt, and so on.
1. When a depends on t
2. When a depends on x or v
a(x): use the chain rule, a = v dv/dx. Then ∫v dv = ∫a(x) dx. Example: a = 2x from rest at x = 1 m gives ½v² = x² − 1.
This connects to energy: it’s the work–energy theorem per unit mass.
a(v): separate variables: dv/f(v) = dt, then integrate both sides. This is how drag problems (Unit 2) lead to exponential solutions.
3. Example, then try it
A particle’s velocity is v(t) = t³ (m/s). Compare its average acceleration from 1 s to 3 s with its instantaneous acceleration at 2 s.
Average: Δv/Δt = (27 − 1) ÷ 2 = 13 m/s².
Instantaneous: a = 3t², so a(2) = 12 m/s². They differ because a isn’t constant.
A particle starts from rest at the origin with a(t) = 6 − 2t (m/s²). Find its maximum velocity and its position at that moment. When does it stop again, and where?
Velocity: v(t) = ∫(6 − 2t) dt = 6t − t² (v0 = 0).
Max v: a = 0 at t = 3 s, so v(3) = 18 − 9 = 9 m/s.
Position: x(t) = 3t² − t³/3, so x(3) = 27 − 9 = 18 m.
Stops again: 6t − t² = 0 at t = 6 s, where x(6) = 108 − 72 = 36 m.
Answer: v max = 9 m/s at 18 m; it stops at t = 6 s, 36 m from the start.
4. Common mistakes
For which motion can you correctly use v = v₀ + at?
On a frictionless incline the acceleration is constant (g sin θ). The other three all have changing accelerations.
Ready to practice?
5 questions on Non-Constant Acceleration, each with a hint and an explanation for every wrong answer.