For a vehicle moving at $80 \, km/hr$, how many times is the stopping distance compared to a vehicle moving at $40 \, km/hr$, assuming both have the same deceleration (in $\times$)?

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.

An object accelerates from rest to a velocity $27.5 \,m/s$ in $10 \,s$. Find the distance covered by the object in the next $10 \,s$ (in $m$).

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

$A$ particle moves along a straight line. Its position at any instant is given by $x = 32t - \frac{8t^3}{3}$,where $x$ is in $m$ and $t$ is in $s$. Find the acceleration of the particle at the instant when the particle is at rest $..........\,m/s^2$.