(D) Given acceleration $a = v \frac{dv}{dx} = -\alpha x^2$.Separating the variables,we get $v dv = -\alpha x^2 dx$.Integrating both sides from the ini

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

$(\frac{3v_0^2}{2\alpha})^{1/3}$

(D) Given acceleration $a = v \frac{dv}{dx} = -\alpha x^2$.Separating the variables,we get $v dv = -\alpha x^2 dx$.Integrating both sides from the initial state $(x=0, v=v_0)$ to the final state $(x=d, v=0)$:$\int_{v_0}^{0} v dv = \int_{0}^{d} -\alpha x^2 dx$.Evaluating the integrals:$[\frac{v^2}{2}]_{v_0}^{0} = -\alpha [\frac{x^3}{3}]_{0}^{d}$.$0 - \frac{v_0^2}{2} = -\alpha (\frac{d^3}{3})$.$\frac{v_0^2}{2} = \frac{\alpha d^3}{3}$.Solving for $d$:$d^3 = \frac{3v_0^2}{2\alpha}$.$d = (\frac{3v_0^2}{2\alpha})^{1/3}$.

Class 11 Physics Motion in a Straight Line Uniformly Accelerated Motion

Difficult

$A$ particle is projected with velocity $v_0$ along the $x-$axis,and its deceleration is given by $a = -\alpha x^2$. The distance at which the particle stops is

A
$\sqrt{\frac{3v_0}{2\alpha}}$
B
$(\frac{3v_0}{2\alpha})^{1/3}$
C
$\sqrt{\frac{3v_0^2}{2\alpha}}$
D
$(\frac{3v_0^2}{2\alpha})^{1/3}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Motion in a Straight Line

Subtopic

Uniformly Accelerated Motion

$A$ particle moving along the $X$-axis has acceleration $f$ at time $t$ given by $f=f_0\left(1-\frac{t}{T}\right)$, where $f_0$ and $T$ are constants. The particle at $t=0$ has zero velocity. In the time interval between $t=0$ and the instant when $f=0$, the particle's velocity is

MediumTS EAMCET 2020

View Solution

$A$ 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 $\,N$)

EasyKCET 2008

View Solution

If a train travelling at $72 \text{ km/h}$ is to be brought to rest in a distance of $200 \text{ m}$,then its retardation should be ............ $\text{m/s}^2$.

Easy

View Solution

$A$ point starts moving in a straight line with a certain acceleration. At a time $t$ after the beginning of motion,the acceleration suddenly becomes a retardation of the same value. The time in which the point returns to the initial point is:

Difficult

View Solution

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

DifficultJEE MAIN 2024

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

$A$ particle is projected with velocity $v_0$ along the $x-$axis,and its deceleration is given by $a = -\alpha x^2$. The distance at which the particle stops is

Similar Questions

$A$ particle moving along the $X$-axis has acceleration $f$ at time $t$ given by $f=f_0\left(1-\frac{t}{T}\right)$, where $f_0$ and $T$ are constants. The particle at $t=0$ has zero velocity. In the time interval between $t=0$ and the instant when $f=0$, the particle's velocity 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 $\,N$)

If a train travelling at $72 \text{ km/h}$ is to be brought to rest in a distance of $200 \text{ m}$,then its retardation should be ............ $\text{m/s}^2$.

$A$ point starts moving in a straight line with a certain acceleration. At a time $t$ after the beginning of motion,the acceleration suddenly becomes a retardation of the same value. The time in which the point returns to the initial point is:

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module