A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = β x⁻²ⁿ, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by (a) -2β² x⁻²ⁿ⁺¹ (b) -2nβ² e⁻⁴ⁿ⁺¹ (c) -2nβ² x⁻²ⁿ⁻¹ (d) -2nβ² x⁻⁴ⁿ⁻¹

Answer: (d). a = vdv/dx dv/dx = -2nβ x⁻²ⁿ⁻¹ a = β x⁻²ⁿ · (-2nβ x⁻²ⁿ⁻¹) = -2nβ² x⁻⁴ⁿ⁻¹ So the answer is (d).

A particle of unit mass undergoes one-dimensional motion such that… | NEET 2015 (AIPMT, Cancelled Paper)

Q34NEET 2015 · AIPMT · CANCELLED PAPERMedium

Velocity, Acceleration & Variable Motion

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v(x) = \beta x^{-2n}$ , where $\beta$ and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

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Answer: (d)

$a = v\dfrac{dv}{dx}$
$\dfrac{dv}{dx} = -2n\beta x^{-2n-1}$
$a = \beta x^{-2n} \cdot (-2n\beta x^{-2n-1}) = -2n\beta^2 x^{-4n-1}$
So the answer is (d).

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Velocity, Acceleration & Variable Motion question from NEET 2015 (AIPMT, Cancelled Paper)

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