Question

Guyzz please answer q.no 21
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Answer 1

Hope u like my process
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Given
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=> x² = 2+ t

=> Acceleration = $\frac{dv}{dt} = \frac{ {d}^{2} x}{ {dt}^{2} } = \frac{d}{dt} ( \frac{dx}{dt} )$

So..
We have to find out dx/dt

Now..

$= > {x}^{2} = 2 + t \\ \\ = > x = \sqrt{2 + t} \\ \\ \underline{ \text{ \bf \: differntiating \: \: both \: \: sides}} \\ \\ = > v = \frac{dx}{dt} = \frac{d( \sqrt{2 + t})}{dt} = \frac{1}{2 \sqrt{2 + t} } \\ \\ \underline{ \bf \:a gain \: \: differentiating \: \: both \: \: sides \: \: w.r.t. \: t \: } \\ \\ = > a = \frac{dv}{dt} = \frac{d}{dt} ( \frac{1}{2 \sqrt{2 + t} } ) \\ \\ \: \: \: \: \: \: \: = \frac{1}{2} \frac{d}{dt} ( \frac{1}{ \sqrt{2 + t} } ) = - \frac{1}{2} \times \frac{1}{2} \times \frac{1}{ { (\sqrt{2 + t} )}^{3} } \\ \\ \: \: \: \: \: \: \: \: \: \: = - \frac{1}{4} \times \frac{1}{ {x}^{ 3} } = \bf \underline{ - \frac{ 1}{4 {x}^{3} } } \text{ \: units}$
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Hope this is ur required answer

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