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  • A particle located at x=0 at time t=0, starts moving along the positive x-direction with a velocity v which varies as <img alt="\mathrm{v=\beta \sqrt{x}}" src="https://entrancecorner.oncodecog

A particle located at x=0 at time t=0, starts moving along the positive x-direction with a velocity v which varies as \mathrm{v=\beta \sqrt{x}}. The velocityof particle varies with time as-                        (\mathrm{\beta } is constant )

Option: 1

\mathrm{t}


Option: 2

\mathrm{\sqrt{t}}


Option: 3

\mathrm{t^2}


Option: 4

\mathrm{\frac{1}{\sqrt{t}}}


Answers (1)

best_answer

Given,

             \mathrm{\begin{aligned} & \quad V=\beta \sqrt{x} \\ & \frac{d x}{d t}=\beta \sqrt{x} \\ & \frac{d x}{\sqrt{x}}=\beta d t \end{aligned}}

Integrate in both sides

\mathrm{\begin{gathered} \int_0^x \frac{d x}{\sqrt{x}}=\int_0^t \beta d t \\ \int_0^x x^{-1 / 2} d x=\int_0^t \beta \cdot d t \\ {\left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_0^x=[\beta t]_0^t} \end{gathered}}

             \mathrm{\begin{aligned} & {\left[\frac{x^{1 / 2}}{1 / 2}\right]_0^x=\beta t} \\ & {[2 \sqrt{x}]_0^x=\beta t} \\ & 2 \sqrt{x}=\beta t \\ & \sqrt{x}=\frac{\beta t}{2} \Rightarrow \end{aligned}}

                 \mathrm{\begin{aligned} & v=\beta \cdot \beta \frac{t}{2} \\ & v=\beta^2 \frac{t}{2} \\ & v\: \alpha \: t \end{aligned}}

 

Posted by

HARSH KANKARIA

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