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Instantaneous acceleration

A particle is...

Question

A particle is moving along a straight-line path according to the relation

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

$S$ represents the displacement covered in $t seconds$ and $a, b, c$ are constants. The acceleration of the particle varies as

S^{- 2 / 3}

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S^{3 / 2}

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S^{2}

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S^{- 3}

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Solution

The correct option is D $S^{- 3}$
According to question

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

Differentiating on both sides,

$2 S \frac{d S}{d t} = 2 a t + 2 b$

$\frac{d S}{d t} = \frac{a t + b}{S}$

Again differentiating w.r.t $t$ , we get

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

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

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

We know that Acceleration, $A = \frac{d^{2} S}{d t^{2}}$

$\Rightarrow A = \frac{a S^{2} - (a t + b)^{2}}{S^{3}}$

$A = \frac{a (a t^{2} + 2 b t + c) - (a t + b)^{2}}{S^{3}} A = \frac{a^{2} t^{2} + 2 a b t + a c - a^{2} t^{2} - b^{2} - 2 a b t}{S^{3}} = \frac{a c - b^{2}}{S^{3}}$

$∴ A \propto S^{- 3}$

Hence, option $(A)$ is the correct answer.

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