NEET Physics MCQs

A particle starts from rest and moves along a straight line with an acceleration given by $a = 6t - 4$, where $a$ is in $m/s^2$ and $t$ is in seconds. At what time is the velocity of the particle zero after the start of motion?

A body travels along a path described by $y = x^2$. What is the ratio of distance to displacement when moving from $(0,0)$ to $(2,4)$?

A particle moves such that its displacement $x$ varies with time $t$ as $x = At^2 + Bt^3$. The graph of velocity versus time will be:

The velocity-time graph of a particle is given by $v = v_0(1-e^{-kt})$, where $v_0$ and $k$ are constants. The displacement of the particle at time $t$ is:

The acceleration of a particle is given by $a = -ω^2x$. The graph representing the variation of velocity ($v$) with displacement ($x$) will be:

A particle starts from rest and moves with a varying acceleration as shown in the graph. If the particle's velocity at $t=4$ s is $v$, its velocity at $t=8$ s will be:

A particle moves in a plane with a velocity given by $\vec{v} = (A\omega cos(\omega t))\hat{i} + (B\omega sin(\omega t))\hat{j}$. If the particle starts from the origin at $t=0$, the equation of the trajectory is:

A particle moves along a path defined by the equation $r = 2t^2\hat{i} + (3t - 1)\hat{j}$, where $r$ is in meters and $t$ is in seconds. At $t = 2$ seconds, a force $F = (4t\hat{i} + 6\hat{j})$ Newtons acts on the particle. What is the instantaneous power delivered by the force at $t = 2$ seconds?

A particle is subjected to a force $F = (3x^2 - 2x + 1)\hat{i}$ N, where $x$ is in meters. What is the work done by this force in moving the particle from $x = 0$ to $x = 2$ m?

A particle moves in one dimension under the influence of a conservative force with potential energy $U(x) = ax^3 - bx$, where $a$ and $b$ are positive constants. The equilibrium position of the particle is at: