$A$ particle moves according to the equation $x = at^2 - bt^3$. At what time will its acceleration be zero?

Two trains travelling on the same track are approaching each other with equal speeds of $40\ m/s$. The drivers of the trains begin to decelerate simultaneously when they are just $2.0\ km$ apart. Assuming the decelerations to be uniform and equal,the value of the deceleration to barely avoid collision should be..........$m/s^2$.

$A$ body starting from rest moves with an acceleration of $\frac{5}{4} \,ms^{-2}$. The distance travelled by the body in the third second is:

$A$ 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 ........... $m/s$.

$A$ train accelerates from rest at a constant rate $\alpha$ for distance $x_1$ and time $t_1$. After that,it retards to rest at a constant rate $\beta$ for distance $x_2$ and time $t_2$. Which of the following relations is correct?

For a particle moving along a straight line path, the displacements in the third and fifth seconds of its motion are $10 \ m$ and $18 \ m$ respectively. The speed of the particle at time $t=4 \ s$ is (in $ms^{-1}$)