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MCQEasyJEE 2025Velocity & Acceleration

JEE Physics 2025 Question with Solution

A particle moves along the xx-axis and has its displacement xx varying with time tt according to the equation x=c0(t2−2)+c(t−2)2x = c_0 (t^2 - 2) + c (t - 2)^2 where c0c_0 and cc are constants of appropriate dimensions. Then, which of the following statements is correct?

  • A

    the acceleration of the particle is 2c02c_0

  • B

    the acceleration of the particle is 2c2c

  • C

    the initial velocity of the particle is 4c4c

  • D

    the acceleration of the particle is 2(c+c0)2(c + c_0)

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The displacement is x=c0(t2−2)+c(t−2)2x = c_0 (t^2 - 2) + c (t - 2)^2.

Find: The correct statement about the particle's motion.

Expand the displacement equation:

x=c0t2−2c0+c(t2−4t+4)x = c_0 t^2 - 2c_0 + c(t^2 - 4t + 4) x=c0t2−2c0+ct2−4ct+4cx = c_0 t^2 - 2c_0 + ct^2 - 4ct + 4c x=(c0+c)t2−4ct+(4c−2c0)x = (c_0 + c)t^2 - 4ct + (4c - 2c_0)

Differentiate with respect to time to get velocity:

v=dxdt=ddt[(c0+c)t2−4ct+(4c−2c0)]v = \frac{dx}{dt} = \frac{d}{dt}\left[(c_0 + c)t^2 - 4ct + (4c - 2c_0)\right] v=2(c0+c)t−4cv = 2(c_0 + c)t - 4c

Differentiate again to get acceleration:

a=dvdt=ddt[2(c0+c)t−4c]a = \frac{dv}{dt} = \frac{d}{dt}\left[2(c_0 + c)t - 4c\right] a=2(c0+c)a = 2(c_0 + c)

Also, the initial velocity is obtained at t=0t = 0:

v0=2(c0+c)(0)−4c=−4cv_0 = 2(c_0 + c)(0) - 4c = -4c

So the statement saying the initial velocity is 4c4c is incorrect.

Therefore, the correct option is D. The acceleration of the particle is 2(c+c0)2(c + c_0).

Direct Differentiation

Given: x=c0(t2−2)+c(t−2)2x = c_0 (t^2 - 2) + c (t - 2)^2

Find: The correct statement among the given options.

Differentiate directly without expanding:

v=dxdt=ddt[c0(t2−2)+c(t−2)2]v = \frac{dx}{dt} = \frac{d}{dt}\left[c_0 (t^2 - 2) + c (t - 2)^2\right] v=c0⋅2t+c⋅2(t−2)v = c_0 \cdot 2t + c \cdot 2(t - 2) v=2c0t+2c(t−2)v = 2c_0 t + 2c(t - 2) v=2c0t+2ct−4cv = 2c_0 t + 2ct - 4c v=(2c0+2c)t−4cv = (2c_0 + 2c)t - 4c

Now differentiate velocity:

a=dvdt=ddt[(2c0+2c)t−4c]a = \frac{dv}{dt} = \frac{d}{dt}\left[(2c_0 + 2c)t - 4c\right] a=2c0+2c=2(c0+c)a = 2c_0 + 2c = 2(c_0 + c)

Check the initial velocity:

v0=(2c0+2c)⋅0−4c=−4cv_0 = (2c_0 + 2c)\cdot 0 - 4c = -4c

Hence the initial velocity is not 4c4c.

Thus, the correct statement is that the acceleration of the particle is 2(c+c0)2(c + c_0). So the correct option is D.

Common mistakes

  • Differentiating c(t−2)2c(t - 2)^2 incorrectly as 2ct2ct. This ignores the shift inside the bracket. Differentiate it as 2c(t−2)2c(t - 2), then simplify.

  • Stopping after finding velocity and comparing options directly. The question asks about acceleration, so differentiate the displacement twice or the velocity once more.

  • Taking the initial velocity as 4c4c instead of −4c-4c. At t=0t = 0, substitute carefully into v=2(c0+c)t−4cv = 2(c_0 + c)t - 4c to get v0=−4cv_0 = -4c.

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