(B) Initial velocity $u = 72 \,km/h = 72 \times \frac{5}{18} \,m/s = 20 \,m/s$.Final velocity $v = 0 \,m/s$ (as the bus comes to a halt).Time $t = 4 \

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) Initial velocity $u = 72 \,km/h = 72 \times \frac{5}{18} \,m/s = 20 \,m/s$.Final velocity $v = 0 \,m/s$ (as the bus comes to a halt).Time $t = 4 \,s$.Using the first equation of motion, $v = u + at$, we find the acceleration:$0 = 20 + a(4) \Rightarrow 4a = -20 \Rightarrow a = -5 \,m/s^2$.The distance $s$ travelled can be calculated using the second equation of motion, $s = ut + \frac{1}{2}at^2$:$s = (20)(4) + \frac{1}{2}(-5)(4)^2$$s = 80 - \frac{1}{2}(5)(16)$$s = 80 - 40 = 40 \,m$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Difficult

$A$ bus moving along a straight highway with a speed of $72 \,km/h$ is brought to a halt within $4 \,s$ after applying the brakes. The distance travelled by the bus during this time (assume the retardation is uniform) is . . . . . . $m$.

JEE MAIN 2024 All JEE MAIN

A
$30$
B
$40$
C
$60$
D
$70$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Motion in Straight Line

Subtopic

Uniformly Accelerated Motion

Previous Year

JEE MAIN 2024 Paper All JEE MAIN years →

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

MediumNEET 2016

View Solution

$A$ particle initially at rest moves along the $x$-axis. Its acceleration varies with time as $a = 4t$. If it starts from the origin,the distance covered by it in $3 \ s$ is $........... \ m$.

Medium

View Solution

The motion of a particle along a straight line is described by the equation $x = 8 + 12t - t^3$,where $x$ is in meters and $t$ is in seconds. The retardation of the particle when its velocity becomes zero is .......... $m/s^2$.

MediumAIPMT 2012

View Solution

What kind of motion is represented by the equation $\Delta S = v \Delta t$?

Easy

View Solution

$A$ car passes three points $A$, $B$ and $C$ at $3 \text{ m/s}$, $6 \text{ m/s}$ and $9 \text{ m/s}$ respectively in a straight line with uniform acceleration. If distance $AB = 60 \text{ m}$, then find the distance $BC$. (in $m$)

DifficultMHT CET 2026

View Solution

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

$A$ bus moving along a straight highway with a speed of $72 \,km/h$ is brought to a halt within $4 \,s$ after applying the brakes. The distance travelled by the bus during this time (assume the retardation is uniform) is . . . . . . $m$.

Similar Questions

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

$A$ particle initially at rest moves along the $x$-axis. Its acceleration varies with time as $a = 4t$. If it starts from the origin,the distance covered by it in $3 \ s$ is $........... \ m$.

The motion of a particle along a straight line is described by the equation $x = 8 + 12t - t^3$,where $x$ is in meters and $t$ is in seconds. The retardation of the particle when its velocity becomes zero is .......... $m/s^2$.

What kind of motion is represented by the equation $\Delta S = v \Delta t$?

$A$ car passes three points $A$, $B$ and $C$ at $3 \text{ m/s}$, $6 \text{ m/s}$ and $9 \text{ m/s}$ respectively in a straight line with uniform acceleration. If distance $AB = 60 \text{ m}$, then find the distance $BC$. (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module