What is the retardation of moving particle if the relation between time ( $t$ ) and position ( $x$ ) is $t = A x^{2} + B x$ , where $A$ and $B$ are constant.

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Solution

The time is given as,

$t = A x^{2} + B x$

Differentiate the above expression with respect to $t$ ,

$1 = 2 A x \frac{d x}{d t} + B \frac{d x}{d t}$

$\frac{d x}{d t} = \frac{1}{(2 A x + B)}$

Again differentiate the above expression with respect to $t$ ,

$\frac{d^{2} x}{d t^{2}} = \frac{- 1}{{(2 A x + B)}^{2}} \times \frac{2 A d x}{d t}$

$a = \frac{- 2 A}{{(2 A x + B)}^{3}}$

Thus, the retardation of the moving particle is $\frac{- 2 A}{{(2 A x + B)}^{3}}$ .

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