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 $a$.

(N/A) Let the distance travelled by the vehicle before it stops be $d_{s}$.
Using the third equation of motion,$v^{2} = v_{0}^{2} + 2ax$,where $v$ is the final velocity,$v_{0}$ is the initial velocity,$a$ is the acceleration (which is negative in this case,i.e.,$-a$),and $x$ is the displacement.
At the point where the vehicle stops,the final velocity $v = 0$.
Substituting these values into the equation: $0^{2} = v_{0}^{2} + 2(-a)d_{s}$.
Rearranging the terms to solve for $d_{s}$:
$2ad_{s} = v_{0}^{2}$
$d_{s} = \frac{v_{0}^{2}}{2a}$.
Thus,the stopping distance is directly proportional to the square of the initial velocity $(d_{s} \propto v_{0}^{2})$.