(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 moving along a straight line with uniform acceleration is given by $V = \sqrt{196 - 16x} \text{ m/s}$, then its acceleration is ($x$ is the displacement of the particle). (in $\text{ m/s}^2$)

MediumAP EAMCET 2023

View Solution

$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

DifficultJEE MAIN 2024

View Solution

The displacement of a particle after time $t$ is given by $x = (k/b^2)(1 - e^{-bt})$ where $b$ is a constant. What is the acceleration of the particle?

Medium

View Solution

$A$ rifle bullet loses $(1/20)^{th}$ of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is

Difficult

View Solution

$A$ particle starts its motion from rest under the action of a constant force. If the distance covered in the first $10 \ s$ is $S_1$ and that covered in the first $20 \ s$ is $S_2$,then:

MediumAIPMT 2009

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 moving along a straight line with uniform acceleration is given by $V = \sqrt{196 - 16x} \text{ m/s}$, then its acceleration is ($x$ is the displacement of the particle). (in $\text{ m/s}^2$)

$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

The displacement of a particle after time $t$ is given by $x = (k/b^2)(1 - e^{-bt})$ where $b$ is a constant. What is the acceleration of the particle?

$A$ rifle bullet loses $(1/20)^{th}$ of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is

$A$ particle starts its motion from rest under the action of a constant force. If the distance covered in the first $10 \ s$ is $S_1$ and that covered in the first $20 \ s$ is $S_2$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module