$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. What is the retardation of the car (assumed uniform in $m/s^2$)?

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

$A$ particle is moving along a straight line with constant acceleration. At the end of the $10^{th}$ second,its velocity becomes $20 \, m/s$,and in the $10^{th}$ second,it travels a distance of $10 \, m$. Then the acceleration of the particle will be ........ $m/s^2$.

$A$ body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n^{\text{th}}$ second of its motion to the total distance travelled in $n$ seconds is

$A$ body starts from rest with uniform acceleration $a$. Its velocity after $n \ s$ is $v$. The displacement of the body in the last $3 \ s$ is: (Assume total time of journey is $n \ s$)

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: