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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 $a$.
Solution
Let the distance travelled by the vehicle before it stops be $d_{s}$. Then, using equation of motion $v^{2}=v_{0}^{2}+2 a x,$ and noting that $v=0,$ we have the stopping distance
$d_{s}=\frac{-v_{0}^{2}}{2 a}$
Thus, the stopping distance is proportional to the square of the inttial velocity. Doubling the iittial velocity increases the stopping distance by a factor of $4$ (for the same deceleration). For the car of a particular make, the braking distance was found to be $10 \mathrm{m}, 20 \mathrm{m}, 34 \mathrm{m}$ and $50 \mathrm{m}$ corresponding to velocities of $11,15,20 $ and $25 \mathrm{m} / \mathrm{s}$ which are nearly consistent with the above formula.
Stopping distance is an important factor considered in setting speed limits, for example. in school zones.
