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Motion in A Straight Line question

2015 · Q135
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Motion in A Straight Line question

2015 · Q135

NEETPhysicsMotion in A Straight LineMCQ+4 / −1
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x)=βx−2nv\left( x \right) = \beta {x^{ - 2n}}v(x)=βx−2n, where β\betaβ 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
  1. A
    −2β2x−2n+1- 2{\beta ^2}{x^{ - 2n + 1}}−2β2x−2n+1
  2. B
    −2nβ2e−4n+1- 2n{\beta ^2}{e^{ - 4n + 1}}−2nβ2e−4n+1
  3. C
    −2nβ2x−2n−1- 2n{\beta ^2}{x^{ - 2n - 1}}−2nβ2x−2n−1
  4. D
    −2nβ2x−4n−1- 2n{\beta ^2}{x^{ - 4n - 1}}−2nβ2x−4n−1
View written solutionFree

Correct answer: D

Given v(x)=βx−2nv\left( x \right) = \beta {x^{ - 2n}}v(x)=βx−2n

∴\therefore∴ dvdx=−2nβx−2n−1{{dv} \over {dx}} = - 2n\beta {x^{ - 2n - 1}}dxdv​=−2nβx−2n−1

So acceleration of the particle is
aaa = vdvdxv{{dv} \over {dx}}vdxdv​ = (βx−2n\beta {x^{ - 2n}}βx−2n) ×\times× (−2nbx−2n−1- 2nb{x^{ - 2n - 1}}−2nbx−2n−1)

              = −2nβ2x−4n−1 - 2n{\beta ^2}{x^{ - 4n - 1}}−2nβ2x−4n−1

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