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  • Need explanation for: - Kinematics - NEET

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to V(x)=bx^{-2n}

where b 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:

  • Option 1)

    -2nb^{2}x^{-4n-1}

  • Option 2)

    -2b^{2}x^{-2n+1}

  • Option 3)

    -2nb^{2}e^{-4n+1}

  • Option 4)

    -2nb^{2}x^{-2n-1}

 

Answers (1)

best_answer

As we learnt in

Acceleration -

Rate of change of velocity with time.

For: a=change in velocity / change in time

a=frac{v}{t}

a=frac{v_{f}-v_{i}}{t}

- wherein

Eg. A car starting from rest accelerates uniformly to a speed of 75 m/s in 12 seconds. What is car’s acceleration ?

v_{i}= 0

v_{f}= 75 m/s

a= ?

a=frac{v_{f}-v_{i}}{t}

=frac{75-0}{12}

a = 6.25 m/s2

 

 

 V(x)=bx^{-2n}\ \Rightarrow a=\frac{dv}{dt}=\frac{dx}{dt}.\frac{dv}{dx}

a=V\frac{dv}{dx}=(bx^{-2n}).(-2n b x^{-2n-1})

a=-2nb^{2}\:x^{-4n-1}


Option 1)

-2nb^{2}x^{-4n-1}

Correct

Option 2)

-2b^{2}x^{-2n+1}

Incorrect

Option 3)

-2nb^{2}e^{-4n+1}

Incorrect

Option 4)

-2nb^{2}x^{-2n-1}

Incorrect

Posted by

Plabita

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