$A$ 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:

$A$ body starts from rest with uniform acceleration. If its velocity after $n^{\text{th}}$ second (last second) is $V$,then its displacement in the last two seconds is

If $x \propto t^{5/2}$,then

Displacement $(x)$ of a particle is related to time $(t)$ as: $x = at + bt^2 - ct^3$,where $a, b$,and $c$ are constants of the motion. The velocity of the particle when its acceleration is zero is given by:

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

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.