Question

please solve and explain..

Answer 1

Hope u like my process
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Given :-
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=> Displacement x in time t is

$= > \bf x = {(t + 5)}^{ - 1}$
Now,
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=> Velocity (v)

$= \bf \frac{dx}{dt} = \frac{d(t + 5) ^{ - 1} }{dt} \\ \\ = \bf \: - \frac{1}{(t + 5) ^{2} }$

=> Acceleration (a)

$= \bf \: \frac{dv}{dt} = \frac{d( - {(t + 5)}^{ - 2} )}{dt} \\ \\ = \bf - ( - 2) \times \frac{1}{ {(t + 5)}^{3} } = \frac{2}{ {(t + 5)}^{3} }$

Thus,

$\bf \: acceleration \: \: is \: \: proportional \: \: to \: \frac{1}{ {(t + 5)}^{3} }$

Now,
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$= > \it \: \frac{1}{ {(t + 5)}^{3} } = \frac{1}{(t + 5) ^{2} } \times \frac{1}{(t + 5)} \\ \\ \bf \: which \: \: is \: \: proportional \: \: to \: \: (v \times \sqrt{v} = {v}^{ \frac{3}{2} } )$

Thus..

We can say that

$\it \: acceleration \: \: is \: \: proportional \: \: to \: \: \bf {(velocity) }^{ \frac{3}{2} }$

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Hence,

Option a(✔️) is the required answer.
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❤️Hope this is ur required answer ♥️

❤️Proud to help you ♥️

Answer 2

Hi,   Please see the attached file!    Thanks!!