In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = α

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Question

In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = αx² + βx. The acceleration of the particle is:

Accepted Answer

Correct Answer: C — −2α/(2αx + β)³Step-by-step Solution:Given: t = αx² + βxDifferentiating with respect to x:dt/dx = 2αx + βVelocity: v = dx/dt = 1/(dt/dx) = 1/(2αx + β)Acceleration: a = v(dv/dx)dv/dx = d/dx [1/(2αx + β)] = −2α/(2αx + β)²a = v × dv/dx = [1/(2αx + β)] × [−2α/(2αx + β)²]a = −2α/(2αx + β)³Why Option A is wrong: This ignores the negative sign and has wrong power of the denominator.Why Option B is wrong: Has wrong coefficient (α instead of 2α) and wrong power in denominator.Why Option D is wrong: Missing the negative sign; acceleration is negative as long as α > 0 (velocity decreasing).Key Concepts from Motion in a Straight Line (NCERT Class 11, Chapter 3):When position-time relation is given, velocity v = dx/dt and acceleration a = dv/dt = v(dv/dx).If t = f(x) is given, then v = 1/(dt/dx) — take the reciprocal of the x-derivative of t.Use the chain rule: a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx). This is a standard technique in kinematics.This type of problem tests calculus-based kinematics, common in NEET and JEE.

In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = αx² + βx. The acceleration of the particle is:

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In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = αx² + βx. The acceleration of the particle is:

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