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Derivative from First Principle

A particle is...

Question

A particle is moving along a straight line path according to the relation :
$s^{2} = a t^{2} + 2 b t + c$
where $s$ represents the distance travelled in $t$ seconds and $a$ , $b$ , $c$ are constants. Then the acceleration of the particle is given as

\frac{a}{s} - \frac{(a t + b)^{2}}{s^{3}}

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\frac{a}{s} - \frac{(a t + b)^{2}}{s^{2}}

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\frac{a t}{s} - \frac{(a t + b)^{2}}{s^{3}}

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\frac{a t}{s} - \frac{(a t + b)^{2}}{s^{2}}

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Solution

The correct option is A $\frac{a}{s} - \frac{(a t + b)^{2}}{s^{3}}$

$s^{2} = a t^{2} + 2 b t + c$

$∴ 2 s \frac{d s}{d t} = 2 a t + 2 b$

or $\frac{d s}{d t} = \frac{a t + b}{s}$ , again differentiating

$\frac{d^{2} s}{d t^{2}} = \frac{a . s - (a t + b) \frac{d s}{d t}}{s^{2}} = \frac{a s - (a t + b) (\frac{a t + b}{s})}{s^{2}}$

$∴ \frac{d^{2} s}{d t^{2}} = \frac{a s^{2} - (a t + b)^{2}}{s^{3}}$

$∴ \frac{d^{2} s}{d t^{2}} = \frac{a}{s} - \frac{(a t + b)^{2}}{s^{3}}$

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