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A JEE kinematics question often starts with a single line: "The position of a particle is r = 3t î + (4t − 5t²) ĵ." Then it asks five things about the motion. Every answer comes from differentiating components, but students lose marks by differentiating the wrong thing. This post works one r(t) all the way through, adds a circular case, and lists the traps.
Where is this in the syllabus?
The 2026-27 reprint of the NCERT Class 11 chapter Motion in a Plane defines the position vector as r = xî + yĵ in Section 3.7.1 and the velocity as the limit of Δr/Δt, v = dr/dt (Eq. 3.28). It notes that the direction of v is along the tangent to the path. The Syllabus for JEE (Main) 2026 lists "motion in a plane" in Unit 2, and the JEE (Advanced) 2026 brochure specifies "kinematics in one and two dimensions (Cartesian coordinates only)". So stick to î and ĵ components. Check the syllabus for your exam year.
Why do vector kinematics questions trip people up?
Shaffer and McDermott (2005) studied university students' ability to treat velocity and acceleration as vectors in one and two dimensions. They identified serious conceptual and reasoning difficulties among introductory students, and some of these were also common among pre-college teachers and physics graduate students. In exam terms, one common version is treating acceleration as "change in speed" rather than change in the velocity vector, which hides the part of the acceleration that only turns the motion.
Worked example 1: one r(t), five answers
r = 3t î + (4t − 5t²) ĵ, in metres, with t in seconds.
Velocity. Differentiate each component: v = 3 î + (4 − 10t) ĵ m/s.
Acceleration. a = dv/dt = −10 ĵ m/s², constant. This is a projectile launched with u_x = 3 m/s and u_y = 4 m/s.
Position and velocity at t = 1 s. r = 3 î − 1 ĵ m and v = 3 î − 6 ĵ m/s, so speed = √(9 + 36) = √45 ≈ 6.71 m/s.
Minimum speed. v_x never changes, so speed is least when v_y = 0, at t = 0.4 s. Then v = 3 î, speed 3 m/s. At that instant v is perpendicular to a.
Direction at t = 1 s. tan θ = −6/3, so v points 63.4° below the +x direction.
How do you split acceleration along and across the path?
Take the unit vector along the velocity, v̂ = v/|v|.
Tangential part: a_t = a · v̂. It changes the speed.
Normal part: a_n = √(|a|² − a_t²). It changes the direction.
Radius of curvature: R = v² / a_n.
Worked example 2: tangential and normal acceleration at t = 1 s
From example 1, a = (0, −10) and v = (3, −6), with |v| = 6.71 m/s.
a · v = 0 × 3 + (−10)(−6) = 60, so a_t = 60 / 6.71 = 8.94 m/s². It is positive, so the particle is speeding up.
a_n = √(100 − 80) = 4.47 m/s².
R = 45 / 4.47 = 10.1 m.
At t = 0.4 s, by contrast, a_t = 0 and all 10 m/s² is normal, so R = 3² / 10 = 0.9 m. The path bends most sharply at the top.
Worked example 3: a circle written as r(t)
r = 2 cos(πt) î + 2 sin(πt) ĵ m.
v = −2π sin(πt) î + 2π cos(πt) ĵ, so speed = 2π ≈ 6.28 m/s, constant.
a = −2π² cos(πt) î − 2π² sin(πt) ĵ = −π² r. Its magnitude is 2π² ≈ 19.7 m/s², directed towards the centre.
r · v = 0 at every instant, so v is perpendicular to r, as it must be on a circle.
Here |r| = 2 m always, so d|r|/dt = 0. Yet the speed is 6.28 m/s. This is the callout above in action.
Check with the circular motion formula: v²/R = (2π)² / 2 = 2π² m/s². The same answer.
Common mistakes and how to avoid them
Differentiating |r|. Always differentiate components, then take the magnitude.
Calling a = d(speed)/dt. That is only the tangential part. In example 3 the speed is constant, but a is 19.7 m/s².
Adding magnitudes of components. Speed is √(v_x² + v_y²), not v_x + v_y.
Using average velocity as average speed. Average velocity is Δr/Δt. Over one full circle in example 3 it is zero; the average speed is 6.28 m/s.
Our post on vector errors in JEE physics covers dot products and components, and the guide to projectile range and height uses the same split as example 1. If the curved path is hard to picture from the equations, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
A practice routine
Take three r(t) expressions, one polynomial, one circular and one mixed, and find v and a for each.
For each, find when v is perpendicular to a, if ever.
Find a_t, a_n and R at one chosen instant, and check a_t² + a_n² = |a|².
For the circular one, confirm a = −ω²r.
Try it now: for r = t² î + 2t ĵ, show that the speed at t = 1 s is √8 m/s and find a_t there.
