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Answer 1

$\huge\mathfrak\pink{hlo \: mate}$

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HOPE U WILL HELP ME

The acceleration of particle is

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

Given is the value of velocity v as a function x,

$\green{v(x)={ bx }^{ -2n }}$

therefore for acceleration to be calculated

$\pink{a=\frac { dv }{ dt } \quad and\quad v=\frac { dx }{ dt }}$

thereby,

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

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

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

Hence,

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

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

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

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

Answer 2

Step-by-step explanation:

The acceleration of particle is

\red{a=-2n\quad b2x^{ -4n-1 }}a=−2nb2x

−4n−1

Given is the value of velocity v as a function x,

\green{v(x)={ bx }^{ -2n }}v(x)=bx

−2n

therefore for acceleration to be calculated

\pink{a=\frac { dv }{ dt } \quad and\quad v=\frac { dx }{ dt }}a=

dt

dv

andv=

dt

dx

thereby,

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

dt

dv

)(

dt

dx

)

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

dx

dv

)xv.

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

dx

dv

=

dx

d(bx

−2n

)

Hence,

\pink{a=(\frac { dv }{ dx } )xv}a=(

dx

dv

)xv

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

−2n−1

)xv

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

−2n−1

)x(bx

−2n

)

\Rightarrow a=-2n\quad b2x^{ -4n-1 }⇒a=−2nb2x

−4n−1