The position of a particle as a function of time $t$ is given by $x(t) = at + bt^2 - ct^3$,where $a, b,$ and $c$ are constants. When the particle attains zero acceleration,its velocity will be:

The acceleration $a$ in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$,where $t$ is the time. If the particle starts with an initial velocity $u = 2\,m/s$ at $t = 0$,then the velocity at the end of $2\,s$ is:

$A$ particle initially at rest is moving along a straight line with an acceleration of $2 \,m/s^2$. At a time of $3 \,s$ after the beginning of motion, the direction of acceleration is reversed. The time from the beginning of the motion in which the particle returns to its initial position is

$A$ particle experiences a constant acceleration for $20 \,s$ after starting from rest. If it travels a distance ${S_1}$ in the first $10 \,s$ and a distance ${S_2}$ in the next $10 \,s$,then:

The stopping distance of a vehicle moving with a given acceleration is proportional to the ....... of its speed.

$A$ body starts with an initial velocity of $10\,m/s$ and moves along a straight line with constant acceleration. When the velocity of the particle is $50\,m/s$,the acceleration is reversed in direction. The velocity of the particle when it again reaches the starting point is $............\,m/s$.