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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 v=\alpha \sqrt{x}, The velocity of particle varies with time as: \alpha is a constant

Option: 1

v \alpha t


Option: 2

v \alpha \sqrt{t}


Option: 3

v \alpha t^2


Option: 4

V \propto \frac{1}{\sqrt{t}}


Answers (1)

best_answer

\text { Given, } v=d \sqrt{x}
\mathrm{\frac{d x}{d t}=\alpha \sqrt{x}}
\frac{d x}{\sqrt{x}}=\alpha d t
Integrate\: it - \int_0^x \frac{d x}{\sqrt{x}}=\int_0^t d d t
\begin{aligned} & \int_0^x x^{-1 / 2} d x=\int_0^t \alpha \cdot d t \\ & {\left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_0^x=[\alpha t]_0^t} \end{aligned}
\left[\frac{x^{1 / 2}}{1 / 2}\right]_0^x=\alpha t
\begin{gathered} {[2 \sqrt{x}]_0^x=\alpha t} \\ 2 \sqrt{x}=\alpha t \\ \sqrt{x}=\frac{\alpha t}{2} \end{gathered}
put the value of \sqrt{x}  in equation (1)
\begin{aligned} & V=\alpha \cdot \frac{\alpha t}{2}=\alpha^2 \frac{t}{2} \\ & V=\alpha t^2 / 2 \\ & V \alpha t \end{aligned}
Ans.

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manish painkra

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