Question

The acceleration vector along x-axis of a particle having initial speed v0 changes with distance as a=√x. What is the distance covered by the particle, when its speed becomes twice that of initial speed?

Answer 1

Given:

The acceleration vector along x-axis of a particle having initial speed v0 changes with distance as a=√x.

To find:

Distance covered when its speed becomes twice that of initial speed.

Calculation:

$\therefore \: acc = a = \sqrt{x}$

$= > v \dfrac{dv}{dx} = \sqrt{x}$

$= > v \: dv = \sqrt{x} \: dx$

Integrating on both sides:

$\displaystyle = > \int \: v \: dv = \int \: \sqrt{x} \: dx$

Putting limits:

$\displaystyle = > \int_{v_{0}}^{2v_{0}} \: v \: dv = \int_{x}^{0} \: \sqrt{x} \: dx$

$\displaystyle = > \bigg \{ \frac{ {v}^{2} }{2} \bigg \}_ {v_{0}}^{2v_{0}} = \dfrac{2 {x}^{ \frac{3}{2} } }{3}$

$= > \dfrac{ 4{v_{0}}^{2} }{2} - \dfrac{ {v_{0}}^{2} }{2}= \dfrac{2 {x}^{ \frac{3}{2} } }{3}$

$= > \dfrac{ 3{v_{0}}^{2} }{2} = \dfrac{2 {x}^{ \frac{3}{2} } }{3}$

$= > \dfrac{ 9{v_{0}}^{2} }{4} = {x}^{ \frac{3}{2} }$

$= > x = { \bigg \{\dfrac{ 9{v_{0}}^{2} }{4} \bigg \}}^{ \frac{2}{3} }$

So, final answer is:

$\boxed{ \red{ \bold{ \large{ x = { \bigg \{\dfrac{ 9{(v_{0})}^{2} }{4} \bigg \}}^{ \frac{2}{3} } }}}}$

Answer 2

The correct answer is $x=(\frac{2}{3}v )^{\frac{4}{3} }$.

Given: a=√x.

To Find: The distance when speed is doubled.

Solution:

a=√x

$\frac{vdv}{dx} =\sqrt{x}$

vdv = √x dx

$\int\limits^a_b {v} \, dv$ = $\int\limits^x_0 {\sqrt{x} } \, dx$ (Here, in the limits a=2v and b= v).

$\frac{4v^{2}-v^{2} }{2} = \frac{2}{3}x^{\frac{3}{2} }$

$\frac{3}{2} v^{2}=\frac{2}{3} x^{\frac{3}{2} }$

$x^{\frac{3}{2} } =(\frac{2}{3}v )^{2}$

$x=(\frac{2}{3}v )^{\frac{4}{3} }$

Hence, the distance covered is $x=(\frac{2}{3}v )^{\frac{4}{3} }$.

#SPJ3