Question

a particle is moves as per the equation v=a√x where a is constant and x is displacement find average velocity in the first s meters of the path​

Answer 1

we know that , a = v(dv/dx)

$\implies \rm v = a\sqrt{x}$

$\implies \rm v = v.\dfrac{dv}{dx}\sqrt{x}$

$\implies \rm 1 = \dfrac{dv}{dx}\sqrt{x}$

$\implies \rm dv = (x)^{\dfrac{-1}{2}} dx$

$\implies\rm \displaystyle \int dv = \displaystyle\int \dfrac{(x)^\dfrac{-1}{2}-1}{\dfrac{1}{2}}$

$\implies\rm\displaystyle \int_0^v dv = 2\displaystyle \int_0^s\sqrt{x}$

$\implies \bf v = 2\sqrt{s}$

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Answer 2

Explanation:

we know that , a = v(dv/dx)

$\implies \rm v = a\sqrt{x}$

$\implies \rm v = v.\dfrac{dv}{dx}\sqrt{x}$

$\implies \rm 1 = \dfrac{dv}{dx}\sqrt{x}$

$\implies \rm dv = (x)^{\dfrac{-1}{2}} dx$

$\implies\rm \displaystyle \int dv = \displaystyle\int \dfrac{(x)^\dfrac{-1}{2}-1}{\dfrac{1}{2}}$

$\implies\rm\displaystyle \int_0^v dv = 2\displaystyle \int_0^s\sqrt{x}$

$\implies \bf v = 2\sqrt{s}$

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