The velocity of a particle is $v = v_{0} + gt + Ft^{2}$. Its position is $x = 0$ at $t = 0$; then its displacement after time $t = 1$ is:

The acceleration of a particle is increasing linearly with time $t$ as $a = 6t$. The particle starts from the origin with an initial velocity $u = 10 \ m/s$. The distance travelled by the particle after $t = 2 \ s$ will be: (in $m$)

$A$ bullet is fired into a fixed target and loses one-third of its velocity after traveling $4 \ cm$. It penetrates a further $D \times 10^{-3} \ m$ before coming to rest. The value of $D$ is:

Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

The speeds of two identical cars are $u$ and $4u$ at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

Derive the equations of motion for constant acceleration using the method of calculus.