Question

# A particle of unit mass undesgro one dimension motion. Such that its verocity varies accordingly to v(x) = $${\beta}\vec{x}^2\pi$$,where $$\beta$$ and n are constant and x is the position of the particle. The accelerationx is the position of the particle. The accelerationof the particle a function of x, is given by

A
2nβ2x2n1
B
2nβ2x4n1
C
2β2x2n+1
D
2nβ2e4n+1

Solution

## The correct option is B $$-2n\beta^2x^{-4n-1}$$The velocity (v) as a function of position of the particle is given by$$v(x)=\beta x^{-2n}$$ ............…(i)The acceleration (a) can be calculated as under:$$a=\dfrac{dv}{dt}=\left(\dfrac{dv}{dt}\right)\left(\dfrac{dx}{dt}\right)=\dfrac{dv}{dx}\times v$$ ............….(ii)Differentiating (i) w.r.t. x, we get,$$\Rightarrow \dfrac{dv}{dx}=-2n\beta x^{-2n-1}$$Putting the above reaction in equation (ii), we get$$\Rightarrow a=(-2n\beta x^{-2n-1})\times (\beta x^{-2n})$$$$\Rightarrow a=-2n\beta^2x^{-4n-1}$$Hence, Option $$B$$ is correct answer.Physics

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