$A$ particle moves in a straight line and its position $x$ at time $t$ is given by $x^2 = 2 + t$. Its acceleration is given by

Particle $A$ is moving along the $x$-axis. At time $t=0$,it has a velocity of $10\,m/s$ and an acceleration of $-4\,m/s^2$. Particle $B$ has a velocity of $20\,m/s$ and an acceleration of $-2\,m/s^2$. Initially,both particles are at the origin. At time $t=2\,s$,the distance between the two particles is $.............\,m$.

$A$ car moving with a velocity $6.25 \,ms^{-1}$ is decelerated with $2.5 \sqrt{v} \,ms^{-2}$ (where $v$ is the instantaneous velocity). The time taken by the car to come to rest is: (in $\,s$)

$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

Brakes are applied in a car moving with a speed of $20\ m/s$,which produces a retardation of $5\ m/s^2$. The distance travelled by the car until it stops is ........ $m$.