If a particle moves along a straight line with the law of motion given by $s^{2} = a t^{2} + 2 b t + c$ . Then, the acceleration varies, are

$\frac{1}{s^{3}}$

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$\frac{1}{s}$

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$\frac{1}{s^{4}}$

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$\frac{1}{s^{2}}$

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Solution

The correct option is A
$\frac{1}{s^{3}}$

Step 1: Given Information
The equation for the motion of particle= $s^{2} = a t^{2} + 2 b t + c$
Step 2: Calculation of the acceleration varies
Find of the acceleration varies.
$s^{2} = a t^{2} + 2 b t + c$
$s = \sqrt{a t^{2} + 2 b t + c}$
Acceleration in terms of displacement is $a = \frac{d^{2} s}{d t^{2}}$
Differentiate both sides with respect to $t$
$\frac{d s}{d t} = (\frac{1}{2}) (\frac{[2 a t + 2 b]}{[\sqrt{a t^{2} + 2 b t + c}]})$
$\frac{d s}{d t} = (\frac{[a t + b]}{[\sqrt{a t^{2} + 2 b t + c}]})$
$\frac{d s}{d t} = (\frac{[a t + b]}{s})$
Step 3: Differentiating, $\frac{d s}{d t}$
Differentiate $\frac{d s}{d t}$ ,
$\frac{d^{2} s}{d t^{2}} = [\frac{\sqrt{(a t^{2} + 2 b t + c)} \frac{d}{d t} (a t + b) - (a t + b) \frac{d}{d t} \sqrt{(a t^{2} + 2 b t + c)}}{s^{2}}]$
$\frac{d^{2} s}{d t^{2}} = [\frac{a \sqrt{(a t^{2} + 2 b t + c)} - (a t + b) \times \frac{(2 a t + 2 b)}{2 \sqrt{(a t^{2} + 2 b t + c)}}}{s^{2}}]$
$\frac{d^{2} s}{d t^{2}} = [\frac{a (a t^{2} + 2 b t + c) - {(a t + b)}^{2}}{s^{3}}]$
$\frac{d^{2} s}{d t^{2}} = [\frac{a (a t^{2} + 2 b t + c) - (a^{2} t^{2} + b^{2} + 2 a b t)}{s^{3}}]$
$\frac{d^{2} s}{d t^{2}} = \frac{[a c - b^{2}]}{s^{3}}$
Therefore,
$\frac{d^{2} s}{d t^{2}} \propto \frac{1}{s^{3}}$
Hence, the correct option is (A).

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