JEE Main 2025 Apr 3 Shift 2, Physics Q14: Motion In One Dimension

JEE Main2025Apr 3, Shift 2Physics

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

$x = c_{0} (t^{2} - 2) + c (t - 2)^{2}$

where $c_{0}$ and $c$ are constants of appropriate dimensions. Then, which of the following statements is correct?

A
the acceleration of the particle is $2 c_{0}$
B
the acceleration of the particle is $2 c$
C
the initial velocity of the particle is $4 c$
D
the acceleration of the particle is $2 (c + c_{0})$

Solution

Differentiate $x (t) = c_{0} (t^{2} - 2) + c (t - 2)^{2}$ to find velocity and acceleration:

$v = \frac{d x}{d t} = 2 c_{0} t + 2 c (t - 2)$ .

$a = \frac{d v}{d t} = 2 c_{0} + 2 c = 2 (c + c_{0})$ .

The acceleration is constant and equal to $2 (c + c_{0})$ , matching option (4). (The initial velocity at $t = 0$ is $v (0) = - 4 c$ , not $4 c$ , so option 3 is incorrect.)

Concept behind this question

Motion In One DimensionNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 3 Shift 2, Physics Q14: Motion In One Dimension

Concept behind this question

More previous year questions on Motion In One Dimension

Practice more Physics