Physics · Motion In A Plane · NEET
Differentiate each component with respect to time separately. v = dr/dt. For r = 4t i + 2t² j, the x-part 4t differentiates to 4, and the y-part 2t² differentiates to 4t. So v = 4 i + 4t j. At t = 1 s, v = 4 i + 4 j. Never differentiate the whole thing as one lump — treat i, j, k parts one at a time because they point in fixed directions.
Differentiate twice. First get velocity components: vx = dx/dt, vy = dy/dt. Then get acceleration components: ax = dvx/dt, ay = dvy/dt. For example x = 5t - 2t², so vx = 5 - 4t and ax = -4 m/s². If y = 10t, then vy = 10 and ay = 0. So acceleration = -4 i + 0 j, magnitude 4 m/s². Notice ax stays -4 at every instant — it does not depend on t.
Because the second derivative of a term like 5t is zero. A linear-in-t coordinate means constant velocity in that direction, and constant velocity means zero acceleration. Acceleration only appears where a coordinate has a t² (or higher) term. This is a fast NEET check: scan each coordinate for a t² — no t² means no acceleration in that direction.
Magnitude: |v| = sqrt(vx² + vy²). Direction (angle with x-axis): tan θ = vy / vx, so θ = tan⁻¹(vy/vx). Example: at t = 1 s, v = 4 i + 4 j, so |v| = sqrt(16 + 16) = 4√2 m/s and θ = tan⁻¹(4/4) = 45° with the x-axis. Always plug in the given time only AFTER differentiating, never before.
Differentiate: v = -ω sinωt i + ω cosωt j. The dot product r·v = -ω sinωt cosωt + ω sinωt cosωt = 0, so v is perpendicular to r. Differentiate again: a = -ω² cosωt i - ω² sinωt j = -ω² r. The minus sign means a points opposite to r, i.e. straight back toward the origin. This is uniform circular motion of radius 1, and a is the centripetal acceleration.
The x and y coordinates of a particle at any time t are x = 5t - 2t² and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is:
The position of a particle is given by r(t) = 4t i + 2t² j + 5 k, where t is in seconds and r in metres. Find the magnitude and direction (with respect to the x-axis) of the velocity v(t) at t = 1 s.
A particle moves so that its position vector is given by r = cosωt x̂ + sinωt ŷ, where ω is a constant. Which of the following is true?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
v = dr/dt (differentiate position once) and a = d²r/dt² (differentiate position twice, or velocity once). Do it separately for each component: vx = dx/dt, vy = dy/dt, vz = dz/dt.
No. In this chapter i, j, k are fixed constant directions, so their derivative is zero. You only differentiate the number in front of each unit vector. (Unit vectors do change in polar/rotating frames, but that is not tested at NEET level here.)
Look for a t² or higher power of t in any coordinate. A term like 2t² gives acceleration; a term linear in t (like 5t) gives constant velocity and zero acceleration; a constant term (like 5) is a fixed offset and contributes nothing to v or a.
They substitute the given time into r(t) first, getting a plain number, and the derivative of a number is zero. The correct order is always: differentiate the full expression first, then put the time value in at the very end.
Here you are given the full r(t) and you find v and a by differentiating. In the constant-acceleration case you are given a fixed a and use the reverse process (v = u + at, r = ut + ½at²). One is calculus forward (differentiate), the other is calculus backward (integrate).
