$A$ vehicle moving with $15 \,km/hr$ comes to rest by covering $5 \,m$ distance by applying brakes. If the same vehicle moves at $45 \,km/hr$, then by applying brakes, it will come to rest by covering a distance of: (in $\,m$)
- A$15$
- B$45$
- C$60$
- D$30$
Explore More
Similar Questions
$A$ particle moves according to the equation $x = at^2 - bt^3$. At what time will its acceleration be zero?
Medium
View Solution$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^3 - 6t^2 + 3t + 4$ meters. The velocity when the acceleration is zero is ........ $m s^{-1}$.
MediumAIPMT 1994
View SolutionThe acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.
Difficult
View SolutionThe distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} \ m/s^2$ in the third second is:
If the velocity of a particle is given by $v = (180 - 16x)^{1/2} \text{ m/s}$,then its acceleration will be ....... $\text{m/s}^2$.
Difficult
View SolutionVedclass 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