$A$ car moving with a velocity of $20 \,m \,s^{-1}$ is stopped in a distance of $40 \,m$. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is (in $\,m$)

Two bodies $A$ and $B$ move in the same straight line starting from the same position. Body $A$ moves with a constant velocity $u$ and body $B$ moves with a constant acceleration $a$ starting from rest. When their velocities become equal,the distance between them is:

$A$ $100\, m$ long train is moving with a uniform velocity of $45\, km/hr$. The time taken by the train to cross a bridge of length $1\, km$ is.........$s$

Two bodies $A$ and $B$ start from the same point at the same instant and move along a straight line. Body $A$ moves with uniform acceleration $a$ and body $B$ moves with uniform velocity $V$. They meet after time $t$. The value of $t$ is

$A$ particle executes the motion described by $x(t) = x_0 (1 - e^{-\gamma t})$ for $t \geqslant 0$ and $x_0 > 0$.
$(a)$ Where does the particle start and with what velocity?
$(b)$ Find the maximum and minimum values of $x(t)$,$v(t)$,and $a(t)$. Show that $x(t)$ increases with time,$v(t)$ decreases with time,and $a(t)$ increases with time.

What is stopping distance for a vehicle? What will be the stopping distance if the initial velocity is doubled?