A particle is projected with velocity v 0 along x - axis . The deceleration on the particle is proportional to the square of the distance from the origin i.e. a=-a x2. The distance at which the particle stops is

Question

A particle is projected with velocity v 0 along x - axis . The deceleration on the particle is proportional to the square of the distance from the origin i.e. a=-a x2. The distance at which the particle stops is (A) √(3 v0/2 a) (B) ((3 v0/2 a))(1/3) (C) √(3 v 02/2 a ) (D) ((3 v02/2 a))(1/3)

Tardigrade Admin · Accepted Answer

a=(d v/d t)=(d v/d x) (d x/d t) =v (d v/d x)=-d x2 or, ∫ limitsV00 v d v=-α ∫ limits0s x2 d x or, [(V2/2)]V00=-α[(n2/3)]0s or,(V02/2)=(d s3/3) ∴ S=[(3 V02/2 d)]

Q. A particle is projected with velocity $v_{0}$ along $x$ - axis . The deceleration on the particle is proportional to the square of the distance from the origin i.e. $a = - a x^{2}$ . The distance at which the particle stops is

$\frac{3 v _{0}}{2 a}$

$(\frac{3 v _{0}}{2 a})^{\frac{1}{3}}$

$\frac{3 v _{0}^{2}}{2 a}$

$(\frac{3 v _{0}^{2}}{2 a})^{\frac{1}{3}}$

$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t}$
$= v \frac{d v}{d x} = - d x^{2}$
or, $V_{0} \int 0 v d v = - α 0 \int s x^{2} d x$
or, $[\frac{V ^{2}}{2}]_{V_{0}}^{0} = - α [\frac{n ^{2}}{3}]_{0}^{s}$
or, $\frac{V _{0}^{2}}{2} = \frac{d s ^{3}}{3}$
$∴ S = [\frac{3 V _{0}^{2}}{2 d}]$

Q. A particle is projected with velocity v0​ along x - axis . The deceleration on the particle is proportional to the square of the distance from the origin i.e. a=−ax2. The distance at which the particle stops is

2a3v0​​​

(2a3v0​​)31​

2a3v02​​​

(2a3v02​​)31​