$A$ vehicle moving with $15 \,km/hr$ comes to rest by covering $5 \,m$ distance by applying brakes. If the same vehicle moves at $45 \,km/hr$, then by applying brakes, it will come to rest by covering a distance of: (in $\,m$)

If the velocity of a body related to displacement $x$ is given by $v = \sqrt{5000 + 24x} \; \text{m/s}$,then the acceleration of the body is $\dots \dots \; \text{m/s}^2$.

The distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as $x^{2}=at^{2}+2bt+c$. If the acceleration of the particle depends on $x$ as $x^{-n}$,where $n$ is an integer,the value of $n$ 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.

If the decrease in velocity is the same in equal intervals of time,then describe the $x-t$ graph.

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 ........ $m$.