MixedJEE Physics · Original learning card10 original chapter questions

Differential and Integral Kinematics

For sufficiently smooth one-dimensional motion, v = dx/dt and a = dv/dt = d²x/dt². Conversely, integrating acceleration and velocity determines v and x up to constants fixed by initial conditions.

Why this shows up in the exam

Polynomial motion laws · Implicit x-t relations · Velocity laws given as functions of time

Learn the idea

Differentiate position to descend to velocity and acceleration; integrate to climb back with initial conditions. Position, velocity, and acceleration are three views of the same motion. Differentiation zooms into rates of change, while integration rebuilds accumulated change and needs a starting value.

🧠 Memory hook: Differentiate down x to v to a; integrate up with starting data.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • v = dx/dt — velocity from position
  • a = dv/dt = d²x/dt² — acceleration from velocity or position
  • x(t) = x0 + integral(v dt) — position from velocity and an initial position

How to approach it

  1. 1Identify the independent variable
  2. 2Differentiate or integrate exactly once per kinematic level
  3. 3Apply initial conditions before numerical substitution

Common slip-ups that cost marks

  • •Dropping integration constants
  • •Confusing distance with signed position
  • •Differentiating an implicit relation as if variables were independent

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?

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