Show that acceleration varies as cube of the distance given

Thread starter chwala
Start date Oct 27, 2023
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Acceleration Calculus distance

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SUMMARY

The discussion confirms that acceleration varies as the cube of the distance, represented mathematically as ##(x^{''}) = 2k^2x^3##. The initial velocity is defined as ##(x^{'}) = kx^2##, where ##k## is a constant. The derivation shows that acceleration is directly proportional to the cube of the distance, establishing a clear relationship between distance and acceleration in this context.

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Oct 27, 2023

chwala

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Homework Statement: See attached.

Relevant Equations: Rate of change

In my approach i have distance as ##(x)## and velocity as ##(x^{'})##, then,

##(x^{'}) = kx^2##

where ##k## is a constant, then acceleration is given by,

##(x^{''}) = 2k(x) (x^{'})##

##(x^{''}) = 2k(x)(kx^2) ##

##(x^{''}) = 2k^2x^3##.

Correct?