JEE Main 2023 · 15 April · morning · Single correct · +4 / −1
The position vector of a particle related to time t is given by
r⃗=(10 t î+15 t² ĵ+7 k̂) m
The direction of net force experienced by the particle is :
- (a) Positive x - axis
- (b) Positive y - axis
- (c) Positive z - axis
- (d) In x - y plane
Answer and solution
Answer (b) Positive y - axis
Solution
Physics sir · 3 steps
Force follows acceleration, so differentiate the position twice and see which component survives.
- Velocity. Differentiate each component once: v⃗ = dr⃗/dt = (10 î + 30t ĵ) m s⁻¹. The constant 7k̂ term vanishes, so the particle never moves along z.
- Acceleration. Differentiate again. The x-velocity 10 is constant, so its derivative is zero. Only the y part changes.
a⃗ = dv⃗/dt = 30 ĵ m s⁻²
- So. By Newton's second law, F⃗ = ma⃗ = 30m ĵ, which points along the positive y-axis. The answer is (b).
The idea to keep. Net force points along acceleration, not along velocity or position. Terms linear in t and constant terms drop out after two derivatives, so only the t² terms set the force.
Answer from the NTA final answer key. Solution by PrepWiser.