Module 3: Kinematics · Theory · Advanced · Prerequisites: K8, Foundations 8 and 9
When the acceleration is not constant, the four equations are gone — and the question becomes what the acceleration depends on.
Mahavakya
Acceleration is not a number. It is a function, and the variable it depends on decides which mathematics will work.
On this page
1. The question to ask first
K8 assumed the acceleration was constant and got four equations out of it. Most real motions do not oblige.
A parachutist meets more air resistance the faster she falls, so her acceleration depends on her velocity. A mass on a spring is pulled harder the further it is stretched, so its acceleration depends on position. A rocket burning fuel at a programmed rate has an acceleration that depends on time.
Three situations, three different dependences — and they need three different mathematical moves. So the first question is never “which equation?” but:
What is the acceleration a function of?
| a(t) | a = dv/dt | integrate over time |
| a(x) | v dv = a dx | integrate velocity against position |
| a(v) | dv/a(v) = dt | separate the variables |
Notice what this table is not. It is not three formulas to remember — every row is the definition of acceleration, rearranged so that the two things you can integrate end up on opposite sides.
2. Case one: a depends on time
The easiest case, because the definition is already in the right form. From a = dv/dt:
v(t) = v0 + ∫ a(t) dt
x(t) = x0 + ∫ v(t) dt
Integrate twice, applying an initial condition at each step. This is exactly K6’s chain — areas walking up from acceleration to velocity to position — done algebraically instead of geometrically.
Worked example
A particle has a(t) = 4 − 2t m/s², with v(0) = 1 m/s and x(0) = 0. Find v(2) and x(2).
v = 1 + 4t − t² → v(2) = 1 + 8 − 4 = 5 m/s
x = t + 2t² − t³/3 → x(2) = 2 + 8 − 8/3 = 7.33 m
Check the degrees before checking the arithmetic. Linear acceleration, quadratic velocity, cubic position — which is what K6 said must happen.
3. Case two: a depends on position
Now the definition is in the wrong form. We have a as a function of x, and dv/dt is a rate with respect to t. Integrating over time would require knowing x(t), which is what we are trying to find.
The chain rule gets us out:
a = dv/dt = (dv/dx)(dx/dt) = v dv/dx
v dv = a dx
Foundation 11 called this the identity most worth memorising, and here is why. Time has been eliminated. The left side involves only velocity, the right only position, and both can now be integrated.
v² = u² + 2 ∫ a(x) dx
Now set a constant and watch what happens. The integral becomes a times the displacement, and the result reduces to v² = u² + 2as.
K8’s third equation is the special case of this one. The general result was always there; constant acceleration merely made the integral easy.
Worked example
A particle moves with a = −kx³, where k > 0. At x = 0 its velocity is v0 in the positive direction. How far does it get before stopping?
No time appears in the question, so use v dv = a dx:
½(v² − v0²) = −k ∫0x x³ dx = −kx⁴/4
v² = v0² − kx⁴/2
It stops where v = 0, so xmax = (2v0²/k)1/4.
Time never entered, and was never needed. That is what makes this identity worth having — questions about where a body gets to should not have to be routed through when it got there.
4. Matching the limits
One detail causes more lost marks in this topic than the calculus does.
When you integrate v dv = a(x) dx as a definite integral, the two sides have different variables — so they need different limits.
∫uv v′ dv′ = ∫x0x a(x′) dx′
Velocities on the left. Positions on the right. Students carry the time limits across from habit — writing the interval from t = 0 to t = 3 on both sides — and the answer that comes out looks entirely plausible.
This is Foundation 9’s substitution rule in a physical setting: when you change the variable, you change the limits. The rule was mathematical there and it is the same rule here. If the two look like separate things to remember, they are not.
5. Case three: a depends on velocity
Now separate the variables the other way. From a(v) = dv/dt:
dv / a(v) = dt
Velocity on one side, time on the other, and both integrable.
Worked example
A body moves through a resisting medium with a = −kv. Its initial velocity is v0. Find v(t).
dv/v = −k dt → ln v = −kt + C
v(t) = v0 e−kt
Exponential decay. The body slows for ever and never quite stops, which is what a resistance proportional to velocity produces — the retarding force fades as fast as the motion does. Notice also that the acceleration is −kv and therefore tends to zero along with the velocity.
The three cases now have a character, not just a method:
a(t) describes something programmed — a motion evolving to a schedule.
a(x) describes geometry — the body is pushed differently depending on where it is. Springs and gravitation.
a(v) describes resistance — the medium fights back in proportion to how fast you go.
6. Reading acceleration off a v²–x graph
Differentiate v² = u² + 2∫a dx with respect to x and the integral disappears:
d(v²)/dx = 2a
So plot v² against x — not v against x — and the slope is twice the acceleration at every point.
Where the curve meets the axis, the particle has stopped.
A straight line means constant acceleration. Curvature means the acceleration varies with position. And the point where the curve meets the horizontal axis is where v² = 0 — the turning point, readable directly.
The companion result is that the area under an a–x graph gives ½Δ(v²) — so Δ(v²) is twice that area. The factor of two is easy to drop and separates a correct answer from an almost-correct one.
Note what this is not. Taking the area under the a–x graph is still doing an integral; it is evaluating one geometrically rather than avoiding calculus. K6’s habit applies here as it did there — ask which integral connects the two axes, and the meaning of the area follows.
7. The single most common error
Given a(t) = 6t, a student writes v = u + at = u + 6t².
Wrong, and instructive. That equation was derived by pulling a constant a out of an integral. Here a is not constant, so it cannot come out, and the correct integration gives v = u + 3t². The answer is out by a factor of two, and nothing about it looks unusual.
Nor can you substitute an average acceleration into a constant-acceleration equation unless the problem justifies it. An average describes the overall change; it does not reproduce the motion at intermediate times, which is exactly K3’s point about averages arriving in a new setting.
The check is one line: is the acceleration constant over the whole interval I am applying this to? If not, go back to a = dv/dt.
Contemplation
Look back at the question this part opened with: what is the acceleration a function of? Constant acceleration is the case where the answer is nothing. It depends on no variable at all, which is why the integrals were trivial and why four equations sufficed.
So K8 was not a separate topic that this one supersedes. It was the simplest answer to the question this part asks, and its four equations were what that answer produces. The mathematics did not get harder here; the acceleration merely stopped being a number.
Asking the question is the skill. The three cases are only what the answers happen to look like — and a fourth kind of dependence, on something we have not met, would need its own move, arrived at the same way.
Common misconceptions
1. v = u + at works if you use a(t)
It does not. The equation exists because a came out of an integral, which requires it to be constant. Integrate properly.
2. An average acceleration can be substituted
It reproduces the overall change and not the motion in between. Only valid where the problem says so.
3. The same limits go on both sides of v dv = a dx
Velocities on the left, positions on the right. Change the variable, change the limits.
4. The constant of integration is algebraic bookkeeping
It is the initial condition — the physical information the differential equation cannot contain. Two bodies with the same acceleration and different starting velocities move differently.
5. The a–x graph lets you avoid calculus
Taking its area is an integral. What you avoid is evaluating one algebraically, which is a convenience rather than an escape.
Reflection
- For a = −kv the body never quite stops. Does it travel a finite distance? Work it out.
- A mass on a spring has a = −ω²x. What shape is its v²–x graph, and what does the shape tell you about the motion?
- Why does the a(x) case give v² rather than v? What would you have to do to get v(t) from it?
Key takeaways
- Ask what the acceleration is a function of before choosing any method.
- a(t): integrate twice over time. a(x): use v dv = a dx. a(v): separate dv/a(v) = dt.
- v² = u² + 2∫a dx is the general result; v² = u² + 2as is its constant-acceleration case.
- When integrating v dv = a dx, velocities go on the left and positions on the right.
- On a v²–x graph the slope is 2a, and the axis crossing is where the body stops.
- The area under an a–x graph is ½Δ(v²). Do not lose the two.
- Constant-acceleration equations cannot be used with a varying acceleration, averaged or otherwise.
What comes next
This part assumed one rule applies throughout. Real journeys change their rule — a train accelerates, then cruises, then brakes, and no single function describes all three. K10 handles motion in stages, where the equations change while the body carries on.
Prerequisites: K8, Foundations 8 and 9 · Reading: 18 min · Practice: 40 min · Difficulty: Advanced