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Constant-acceleration formulas cover most of Class 11 kinematics. Then a question says "a particle moves so that its retardation is proportional to its speed" and v = u + at quietly stops working, because a is no longer constant. This post shows you how to choose the right form of acceleration, separate variables and integrate, using three worked examples, and how to check each answer.
Is this in the syllabus?
The CBSE Physics curriculum for 2026-27 lists "Elementary concepts of differentiation and integration for describing motion" and the "calculus treatment" of uniformly accelerated motion in Chapter 2. Chapter 2 of the NCERT Class 11 Physics textbook (2026-27 reprint) derives the constant-acceleration equations by integration and writes a = dv/dt = (dv/dx)(dx/dt) = v dv/dx, "or, v dv = a dx". The JEE (Main) 2026 syllabus lists instantaneous velocity and non-uniform motion in its kinematics unit, and integration and differential equations in mathematics. The NEET (UG) 2026 kinematics unit has similar wording, but integrals of this kind turn up mostly in JEE-style problems. If you need a refresher on the calculus itself, start with our calculus for JEE physics guide.
Which form of acceleration should you use?
Look at what acceleration depends on, then at what the question asks for.
a given as a function of t: v = u + ∫a dt, then x = ∫v dt.
a given as a function of v, question asks about time: use a = dv/dt, so dt = dv / a(v).
a given as a function of v, question asks about distance: use a = v dv/dx, so dx = v dv / a(v).
a given as a function of x: use v dv = a(x) dx and integrate both sides.
The rule is to keep only two variables in the equation, one on each side.
Worked example 1: retardation proportional to speed
A boat's engine cuts out when it is moving at 20 m/s. Water resistance then gives it a = −kv with k = 0.5 s⁻¹. How far does it travel before stopping, and how long does it take to slow to 5 m/s?
Distance. Use v dv/dx = −kv. Cancel v: dv/dx = −k, so v = 20 − 0.5x. Speed falls linearly with distance. It reaches zero at x = 20/0.5 = 40 m. At x = 30 m, v = 20 − 15 = 5 m/s.
Time. Use dv/dt = −kv, so dv/v = −k dt. Integrate from 20 to v: ln(v/20) = −0.5t, so v = 20e^(−0.5t). For v = 5 m/s, t = ln 4 / 0.5 = 2.77 s.
Notice that speed falls exponentially with time but never reaches zero in a finite time, yet the total distance is finite, 40 m. That is a common JEE true-or-false statement.
Worked example 2: retardation proportional to speed squared
A body moving at 30 m/s experiences a = −kv², with k = 0.01 m⁻¹. Find the time and the distance taken to halve its speed.
Time. dv/v² = −k dt. Integrate: 1/v − 1/30 = 0.01t. At v = 15 m/s: 1/15 − 1/30 = 1/30 = 0.01t, so t = 3.33 s. In general, t = 1/(k v₀) for halving the speed.
Distance. v dv/dx = −kv², so dv/v = −k dx. Integrate: ln(v/30) = −0.01x. At v = 15 m/s, x = ln 2 / 0.01 = 69.3 m.
Check the units: k v has units of 1/s, matching the 1/(k v₀) time, and 1/k is in metres, matching the distance.
Worked example 3: acceleration as a function of position
A particle moves along x with a = 3x² (a in m/s², x in m). It passes x = 0 with speed 2 m/s. Find its speed at x = 2 m.
Use v dv = 3x² dx. Integrate from (x = 0, v = 2) to (x = 2, v): (v² − 4)/2 = x³ evaluated from 0 to 2 = 8. So v² = 4 + 16 = 20 and v = 4.47 m/s.
Now a trap: if the particle started at rest at x = 0, the same equation gives v = 0 for ever. At x = 0 the acceleration is zero, so a particle at rest there never moves. The integral is right; it describes an equilibrium point.
We checked all three answers by stepping the motion forward numerically in small time steps. They agree with the integrals to three significant figures, which is a check you can do yourself in a spreadsheet.
Why do students get stuck on these?
Usually not because the integrals are hard. A study by Amos and Heckler (2015) notes evidence that "a major obstacle to student success in the construction of physics integrals is an inability to formulate and interpret differentials and products involving differentials." Writing v dv = a dx is exactly that kind of step, so practise it in words: "a small change in speed, times speed, equals acceleration times a small step in position."
Common mistakes:
Using v² = u² + 2as with a variable a. It only holds for constant acceleration.
Separating with three variables left. If a depends on v and you want x, use v dv/dx, not dv/dt.
Forgetting limits. Put the starting values in as the lower limits instead of adding a constant and then forgetting it.
Losing the sign of retardation. Write a = −kv, not a = kv, for a body slowing down.
A practice routine
For each of the four forms above, write one problem of your own and solve it with limits.
Rework example 1 with k = 0.25 s⁻¹. Predict first: the stopping distance should double to 80 m.
For every answer, check one limiting case, such as k → 0, where the body should move at constant speed.
Mix these with motion at constant power problems, where force also depends on speed.
When you cannot see which form to use, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, solve example 2 again for a third of the starting speed, then check your time against t = 2/(k v₀).
