JEE PYQ: Motion in a Straight Line - Question ID 0266e2d8a4b8 (JEE Main 2025)

When a particle moves along a straight line, its position $x(t)$ is given as a function of time. The velocity $v(t)$ is the first derivative of $x(t)$ with respect to time $t$, and the acceleration $a(t)$ is the first derivative of $v(t)$ (or the second derivative of $x(t)$). Mathematically:

• $v(t) = \frac{dx}{dt}$
• $a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

The question provides the displacement function: $x(t) = c_0(t^2 - 2) + c(t - 2)^2$ Our goal is to compute the acceleration $a(t)$ and the initial velocity $v(0)$, then match them with the given options.

Step 1: Expand the given displacement function

First, expand $x(t)$ to simplify differentiation: $x(t) = c_0(t^2 - 2) + c(t - 2)^2$ $= c_0 t^2 - 2 c_0 + c(t^2 - 4t + 4)$ $= c_0 t^2 - 2 c_0 + c t^2 - 4 c t + 4 c$ Combine like terms: $x(t) = (c_0 + c) t^2 - 4 c t + (4 c - 2 c_0)$

Step 2: Compute the velocity $v(t)$

Velocity is the first derivative of $x(t)$: $v(t) = \frac{dx}{dt} = \frac{d}{dt} \left[ (c_0 + c) t^2 - 4 c t + (4 c - 2 c_0) \right]$ $= 2(c_0 + c) t - 4 c$

Step 3: Compute the acceleration $a(t)$

Acceleration is the first derivative of $v(t)$: $a(t) = \frac{dv}{dt} = \frac{d}{dt} \left[ 2(c_0 + c) t - 4 c \right]$ $= 2(c_0 + c)$ This shows that the acceleration is constant and equal to $2(c_0 + c)$.

Step 4: Compute the initial velocity $v(0)$

Substitute $t = 0$ into the velocity expression: $v(0) = 2(c_0 + c)(0) - 4 c = -4 c$ This value does not match option D ($4 c$), so option D is incorrect.

Step 5: Match with the given options

From Step 3, the acceleration is $2(c_0 + c)$, which matches Option A.

Trap 1: Students often forget to expand $(t - 2)^2$ and attempt to differentiate directly. This can lead to errors in applying the chain rule incorrectly.
Trap 2: Some students confuse the constants $c_0$ and $c$, leading them to select options B or C, which only account for one constant.
Trap 3: Option D mentions the initial velocity as $4 c$. Students might miscalculate $v(0)$ as $+4 c$ instead of $-4 c$, especially if they rush through the algebra.
Exam Tip: Always expand polynomial expressions before differentiating. Double-check signs when evaluating at $t = 0$. Acceleration is the second derivative of displacement, so ensure you compute it correctly.