Question

A particle of unit mass undesgro one dimension motion. Such that its verocity varies accordingly to v(x) = $\beta {\overrightarrow{x}}^2\pi$,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. $-2n{\beta }^2{x}^{-}2n-1$
• B. $-2n{\beta }^2{x}^{-4n-1}$
• C. $-2{\beta }^2{x}^{-2n+1}$
• D. $-2n{\beta }^2{e}^{-}4n+1$

Answer 1

The correct option is B $-2n{\beta }^2{x}^{-4n-1}$

The velocity (v) as a function of position of the particle is given by

$v\left(x\right)=\beta x^{-2n}$ ............…(i)

The acceleration (a) can be calculated as under:

$a=\frac{dv}{dt}=\left(\frac{dv}{dt}\right)\left(\frac{dx}{dt}\right)=\frac{dv}{dx}\times v$ ............….(ii)

Differentiating (i) w.r.t. x, we get,

$\Rightarrow \frac{dv}{dx}=-2n\beta x^{-2n-1}$

Putting the above reaction in equation (ii), we get

$\Rightarrow a=(-2n\beta x^{-2n-1})\times \left(\beta x^{-2n}\right)$

$\Rightarrow a=-2n{\beta }^2{x}^{-4n-1}$

Hence, Option $B$ is correct answer.