Consider a vehicle moving with a velocity $54 \,km/h$. At a distance of $400 \,m$ from the traffic light, brakes are applied. The acceleration of the vehicle after the application of brakes is $-0.3 \,m/s^2$. The vehicle's position relative to the traffic light is: (in $\,m$)
- A$25$
- B$375$
- C$425$
- D$30$
Explore More
Similar Questions
If average and instantaneous velocities are same,then what is the type of motion?
Medium
View SolutionThe acceleration $a$ in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$,where $t$ is the time in seconds. If the particle starts with an initial velocity $v = 2\,m/s$ at $t = 0$,then the velocity at the end of $t = 2\,s$ is $............m/s$.
Medium
View SolutionWhen is the average velocity of an object equal to its instantaneous velocity?
Medium
View SolutionAn object moving along the $x$-axis with a uniform acceleration has a velocity $\vec{v} = (12 \ cm \ s^{-1}) \hat{i}$ at $x = 3 \ cm$. After $2 \ s$,if it is at $x = -5 \ cm$,then its acceleration is:
$A$ body of mass $4 \,kg$ is accelerated by a constant force. It travels a distance of $5 \,m$ in the first second and a distance of $2 \,m$ in the third second. The force acting on the body is: (in $\,N$)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo