$A$ particle starting from rest and moving with constant acceleration travels a distance $x$ in the first $2$ seconds and a distance $y$ in the next $2$ seconds. Then:

$A$ particle moving along the $x-$axis has acceleration $f$ at time $t$,given by $f = f_0(1 - t/T)$,where $f_0$ and $T$ are constants. The particle at $t = 0$ has zero velocity. In the time interval between $t = 0$ and the instant when $f = 0$,the particle's velocity $(v_x)$ is:

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?

$A$ car moving with uniform acceleration covers a distance of $200 \,m$ in the first $2 \,s$ and a distance of $220 \,m$ in the next $4 \,s$. The velocity of the car after $7 \,s$ is: (in $\,m/s$)

$A$ 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 $............\,m/s$.

The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $\omega = a \sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $v_0$, the distance the car will travel and the time it takes before it stops are: