Maths › Mechanics › Kinematics with variable acceleration
Kinematics with variable acceleration
When acceleration refuses to stay constant, suvat retires and calculus takes over: differentiate down the ladder from displacement to acceleration, integrate back up, and let boundary conditions pin the constants.
Builds on Differentiating powers of x and Integration as antidifferentiation.
IN THIS TOPIC
- Differentiate x(t) to get velocity and acceleration; integrate a(t) back with constants found from conditions.
- Find times and positions where a particle is at rest or changes direction.
- Distinguish displacement from total distance when the motion reverses.
WHAT YOU PROBABLY THINK
If a particle's velocity is zero at some instant, its acceleration is zero then too.
The calculus ladder
Velocity is the rate of change of displacement, acceleration the rate of change of velocity. Going down the ladder is differentiation; coming back up is integration, and every integral brings a constant that only a known condition, “starts at the origin”, “initially at rest”, can fix. Constant acceleration is the special case where the ladder reproduces suvat exactly.
Reading the motion
“At rest” means v = 0, and solving it locates the turning moments of the motion. Between those moments the sign of v says which way the particle moves, and total distance must be added leg by leg, while displacement is allowed to cancel.
WORKED EXAMPLE
One particle, fully read
A particle moves with v = 6t − 3t² m s⁻¹. Find its acceleration at t = 1.5, and its displacement from t = 0 to t = 2.
Differentiate: a = 6 − 6t, so at t = 1.5, a = 6 − 9 = −3 m s⁻²: already slowing.
Integrate: x = 3t² − t³ (+ 0, starting at the origin). At t = 2, x = 12 − 8 = 4 m.
The check: v = 3t(2 − t) is zero at t = 0 and t = 2, so the particle moved forward the whole time and 4 m is both displacement and distance.
THE EXAM BIT
- The words are triggers: “at rest” means v = 0, “velocity is constant” means a = 0, “returns to the start” means x = 0.
- Every integration needs its constant, and the constant needs a stated condition.
- For distance when the motion reverses, split the time interval at v = 0 and add magnitudes.
- Keep units on final answers; the calculus will not carry them for you.
CHECK YOURSELF
A particle starts from rest and has a = 12t − 6. Find v(t), and the time after t = 0 at which it is next at rest.
Show a hint
Integrate once; the constant is fixed by starting from rest.
Show the answer
v = 6t² − 6t + c, and v(0) = 0 gives c = 0, so v = 6t² − 6t.
v = 6t(t − 1) = 0 at t = 1 s.
Differentiate x to v to a; integrate back and let the conditions fix the constants.
Solve v = 0 to find the turning moments, and add distance leg by leg.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Differentiate x(t) to get velocity and acceleration; integrate a(t) back with constants found from conditions.
- Find times and positions where a particle is at rest or changes direction.
- Distinguish displacement from total distance when the motion reverses.
No animated video for this topic yet; these notes stand alone.