Question 138: Motion in 1D

Question

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $$v\left( x \right) = \beta {x^{ - 2n}}$$, 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

Accepted Answer

Given $$v\left( x ight) = \beta {x^{ - 2n}}$$
$$ 	herefore $$ $${{dv} \over {dx}} =  - 2n\beta {x^{ - 2n - 1}}$$
So acceleration of the particle is
$$a$$ = $$v{{dv} \over {dx}}$$ = ($$\beta {x^{ - 2n}}$$) $$ 	imes $$ ($$- 2nb{x^{ - 2n - 1}}$$)
= $$ - 2n{\beta ^2}{x^{ - 4n - 1}}$$

Answer

$$ - 2{\beta ^2}{x^{ - 2n + 1}}$$

Answer

$$ - 2n{\beta ^2}{e^{ - 4n + 1}}$$

Answer

$$ - 2n{\beta ^2}{x^{ - 2n - 1}}$$

Answer

$$ - 2n{\beta ^2}{x^{ - 4n - 1}}$$

Motion in 1D

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v\left( x \right) = \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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