Question

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

Answer 1

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Answer 2

Answer: The acceleration of particle is $a=-2n\quad b2x^{ -4n-1 }$

Given is the value of velocity v as a function x, $v(x)={ bx }^{ -2n }$

$\Rightarrow$ therefore for acceleration to be calculated $a=\frac { dv }{ dt } \quad and\quad v=\frac { dx }{ dt } ,$

thereby,

             $a=(\frac { dv }{ dt } )(\frac { dx }{ dt } )$

             $\Rightarrow a=(\frac { dv }{ dx } )xv.$

Therefore, $\frac { dv }{ dx } =\frac { d(bx^{ -2n }) }{ dx }$

             $\Rightarrow \frac { dv }{ dx } =-2n\quad bx^{ -2n–1 }.$

Hence, $a=(\frac { dv }{ dx } )xv$

             $\Rightarrow a=(-2n\quad bx^{ -2n-1 })xv$

             $\Rightarrow a=(-2n\quad bx^{ -2n-1 })x(bx^{ -2n })$

             $\Rightarrow a=-2n\quad b2x^{ -4n-1 }$