$A$ bullet fired into a target loses one-third of its velocity after travelling a distance $x$ meters into the target. If the bullet comes to rest by travelling a further distance $x^{\prime}$, then the ratio $\frac{x^{\prime}}{x}$ is

$A$ moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is

$A$ motorist starting a car from rest accelerates uniformly to a speed of $v \ m/s$ in $9 \ s$. He maintains this speed for another $50 \ s$ and then applies the brakes and decelerates uniformly to rest. His deceleration is numerically equal to three times his previous acceleration. Then the time during which the deceleration takes place is .......... $s$ :-

$A$ particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15) \ m$. The velocity of the body when its acceleration becomes zero is ........... $m/s$.

If a particle moves according to the equation $v = at$,what distance will it cover in the first $4 \, s$ (in $a$)?

The average velocity of a body moving with uniform acceleration travelling a distance of $3.06 \ m$ is $0.34 \ ms^{-1}$. If the change in velocity of the body is $0.18 \ ms^{-1}$ during this time,its uniform acceleration is ......... $ms^{-2}$.