The velocity $v$ of a particle as a function of its position $x$ is expressed as $v = \sqrt{c_1 - c_2 x}$,where $c_1$ and $c_2$ are positive constants. The acceleration of the particle is

$A$ particle is moving along the $X$-axis with velocity $v = e^{-\beta x}$. At time $t = 0$,the particle is located at $x = 0$. The displacement of the particle as a function of time is

On what factors does the stopping distance depend?

Particle $A$ is moving along the $x$-axis. At time $t=0$,it has a velocity of $10\,m/s$ and an acceleration of $-4\,m/s^2$. Particle $B$ has a velocity of $20\,m/s$ and an acceleration of $-2\,m/s^2$. Initially,both particles are at the origin. At time $t=2\,s$,the distance between the two particles is $.............\,m$.

The displacement $x$ (in $m$) of a particle of mass $m$ (in $kg$) moving in one dimension under the action of a force,is related to time $t$ (in $s$) by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero,will be

The distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as $x^{2}=at^{2}+2bt+c$. If the acceleration of the particle depends on $x$ as $x^{-n}$,where $n$ is an integer,the value of $n$ is