If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

$A$ car is moving with a speed of $150 \ km/h$ and after applying the brakes,it travels $27 \ m$ before it stops. If the same car is moving with a speed of one-third of the initial speed,then it will stop after traveling how many meters?

$A$ particle of mass $m$ is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example,when the particle is to the left of $E$,$F$ points to the right. The magnitude of $F$ is constant except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards the left from the equilibrium position $E$ and released from rest at $t=0$. Find the minimum time it will take to reach from $x=-A/2$ to $x=0$.

The deceleration experienced by a moving motor boat,after its engine is cut-off,is given by $\frac{dv}{dt} = -kv^3$,where $k$ is a constant. If $v_0$ is the magnitude of the velocity at cut-off,the magnitude of the velocity at a time $t$ after the cut-off is:

Two particles $A$ and $B$ start from rest and move for equal time on a straight line. Particle $A$ has an acceleration of $2\,m/s^2$ for the first half of the total time and $4\,m/s^2$ for the second half. The particle $B$ has acceleration $4\,m/s^2$ for the first half and $2\,m/s^2$ for the second half. Which particle has covered a larger distance?

The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$, then at $t = 2 \ s$, its position is: (in $m$)