What is stopping distance?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Stopping distance is the distance a vehicle travels after the brakes are applied until it comes to a complete stop.
It depends on the initial velocity $(v_0)$ of the vehicle and the deceleration $(a)$ produced by the brakes.
Using the third equation of motion,$v^2 = v_0^2 + 2as$,where $v$ is the final velocity $(0 \ m/s)$,$v_0$ is the initial velocity,$a$ is the negative acceleration (deceleration),and $s$ is the stopping distance.
Setting $v = 0$,we get $0 = v_0^2 - 2as$,which simplifies to $s = \frac{v_0^2}{2a}$.

Explore More

Similar Questions

Consider that a truck is moving initially with $54 \,km/h$. It is stopped by the driver after looking at an obstacle with a deceleration of $10 \,m/s^2$. The distance travelled by the truck before coming to rest is (in $\,m$)

Derive the equations of uniformly accelerated motion by the graphical method.

Difficult
View Solution

The $v^2$ versus $s$ graph of a particle moving in a straight line is shown in the figure. From the graph,some conclusions are drawn. State which statement is wrong?

$A$ particle travels $10\,m$ in the first $5\,s$ and $10\,m$ in the next $3\,s$. Assuming constant acceleration,what is the distance travelled in the next $2\,s$ (in $m$)?

Difficult
View Solution

$A$ bullet fired into a target loses one-third of its velocity after travelling a distance $x$ meters into the target. If the bullet comes to rest by travelling a further distance $x^{\prime}$, then the ratio $\frac{x^{\prime}}{x}$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo