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Kinematics with variable acceleration
When acceleration is not constant, the suvat equations no longer apply and calculus takes over. Differentiate down the ladder from displacement to acceleration, integrate back up, and let the 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, and integrate a(t) back with constants found from stated conditions.
- Find times and positions where a particle is at rest or changes direction.
- Distinguish displacement from total distance when the motion reverses.
- Differentiate and integrate a position vector in i and j notation, and find speed as the magnitude of the velocity.
COMMON MISCONCEPTION
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 and 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 stated condition can fix, something like “starts at the origin” or “initially at rest”. Constant acceleration is the special case where this ladder reproduces suvat exactly.
Reading the motion
“At rest” means v = 0, and solving that locates the moments where the motion turns. Between those moments the sign of v says which way the particle is going. Total distance then has to be added leg by leg as magnitudes, 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.
Check: v = 3t(2 − t) is zero at t = 0 and t = 2, so the particle moved forward throughout and 4 m is both the displacement and the distance.
The same ladder, in two dimensions
Nothing changes when the motion leaves the line. Write the position as r = x(t)i + y(t)j and differentiate each component separately to get v, then again to get a. Integrating works the same way, one component at a time, and the constant of integration is itself a vector.
Speed is the magnitude of the velocity, so it needs Pythagoras after the differentiation. If r = (3t² − t)i + (t³ + 2)j metres, then v = (6t − 1)i + 3t²j, and at t = 2 the velocity is 11i + 12j with speed √(11² + 12²) = √265 ≈ 16.3 m s⁻¹.
Two phrasings are worth decoding in advance. “Moving parallel to i” means the j component of the velocity is zero. “Moving north-east” means the i and j components are equal and both positive. Set the right component to zero and the rest is algebra you already own.
ASSESSMENT FOCUS
- 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. Write the condition down before you use it.
- For distance when the motion reverses, split the interval at v = 0 and add the magnitudes of the pieces.
- In vector questions, differentiate the components and only take the magnitude at the end. Taking it early loses the direction information the next part wants.
- 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.
Or read them with their worked answers on the kinematics with variable acceleration questions page.
CHECK YOUR PROGRESS
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- Differentiate x(t) to get velocity and acceleration, and integrate a(t) back with constants found from stated conditions.
- Find times and positions where a particle is at rest or changes direction.
- Distinguish displacement from total distance when the motion reverses.
- Differentiate and integrate a position vector in i and j notation, and find speed as the magnitude of the velocity.
Open the full revision checklist to see every objective in the course in one place.