$A$ particle moving with uniform acceleration travels $24\, m$ and $64\, m$ in the first two consecutive intervals of $4\, s$ each. Its initial velocity is ....... $m/s$.

$A$ body is moving along a straight line with initial velocity $v_0$. Its acceleration $a$ is constant. After $t$ seconds,its velocity becomes $v$. The average velocity of the body over the given time interval is

$A$ 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 $\,m/s$)

$A$ car moving at a speed of $30 \, km/hr$ comes to a stop after traveling $8 \, m$ when the brakes are applied. If the same car is moving at a speed of $60 \, km/hr$,what distance (in $m$) will it travel before coming to a stop after the brakes are applied?

$A$ body starts from rest. What is the ratio of the distance travelled by the body during the $4^{th}$ and $3^{rd}$ second?

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: