The velocity of an object is given by $v = kt$,where $k = 2 \, m/s^2$. What distance (in $m$) will it cover in the first $3 \, s$?

Object $A$ starts from rest with a constant acceleration $a$. Object $B$ starts from the same position and moves in the same direction as $A$ with a constant velocity $v$. If both meet after time $t$,then $t =$

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

Tripling the speed of the motor car multiplies the distance needed for stopping it by

The acceleration $a$ (in $m/s^2$) of a body,starting from rest,varies with time $t$ (in seconds) according to the relation $a = 3t + 4$. The velocity of the body at time $t = 2 \ s$ will be $........ \ m/s$.

The speed of a body moving with uniform acceleration is $u$. This speed is doubled while covering a distance $S$. When it covers an additional distance $S$,its speed would become