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 $a$.
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.
Starting from rest, acceleration of a particle is $a = 2(t - 1).$ The velocity of the particle at $t = 5\,s$ is.........$m/sec$
Draw $x \to t$ graph for negative acceleration.
Figure gives the $x -t$ plot of a particle executing one-dimensional simple harmontc motion. Give the signs of position, velocity and acceleration variables of the particle at $t=0.3 \;s , 1.2\; s ,-1.2\; s$