The velocity of a bullet is reduced from $200\,m/s$ to $100\,m/s$ while travelling through a wooden block of thickness $10\,cm$. Assuming the retardation to be uniform,the retardation will be:

An object moves in a straight line with deceleration whose magnitude varies with velocity as $3 v^{2/3}$. If at an initial point,the velocity is $8 \,m/s$,then the distance travelled by the object before it stops is (in $\,m$)

$A$ bullet loses $1/20$ of its velocity after passing through a wooden plank. What is the minimum number of such planks required to stop the bullet completely?

$A$ particle starts from rest at $x=0 \, m$ with an acceleration of $1 \, m/s^2$. At $t = 5 \, s$,it receives an additional acceleration in the same direction as its motion. At $t = 10 \, s$,its speed and position are $v$ and $x$,respectively. Had the additional acceleration not been provided,its speed and position would have been $v_0$ and $x_0$,respectively. It is found that $x - x_0 = 12.5 \, m$. Then one can conclude that $v - v_0$ is .............. $m/s$.

What is stopping distance?

$A$ particle starts moving rectilinearly at time $t = 0$ such that its velocity $v$ changes with time $t$ according to the equation $v = t^2 - t$,where $t$ is in seconds and $v$ is in $m/s$. The time interval for which the particle retards is