JEE MAIN - Physics (2025 - 3rd April Evening Shift - No. 3)

A particle moves along the $x$-axis and has its displacement $x$ varying with time t according to the equation:

$$ x=\mathrm{c}_0\left(\mathrm{t}^2-2\right)+\mathrm{c}(\mathrm{t}-2)^2 $$

where $\mathrm{c}_0$ and c are constants of appropriate dimensions.

Then, which of the following statements is correct?

the acceleration of the particle is $2\left(c+c_0\right)$

the acceleration of the particle is $2 c_0$

the acceleration of the particle is 2 c

the initial velocity of the particle is $4 c$

To determine the acceleration of the particle, we start by differentiating the displacement function with respect to time $ t $.

The displacement of the particle is given by:

$ x = c_0(t^2 - 2) + c(t - 2)^2 $

First, compute the velocity $ v $ by differentiating $ x $ with respect to $ t $:

$ v = \frac{dx}{dt} = \frac{d}{dt}[c_0(t^2 - 2) + c(t - 2)^2] $

This gives:

$ v = 2c_0 t + 2c(t - 2) $

Next, determine the acceleration $ a $ by differentiating the velocity function with respect to time $ t $:

$ a = \frac{dv}{dt} = \frac{d}{dt}[2c_0 t + 2c(t - 2)] $

Calculating this gives:

$ a = 2c_0 + 2c $

Hence, the acceleration of the particle is $ 2c_0 + 2c $.