Question

solve all the questions with explanation.

Answer 1

Hope u like my process
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To find a point of maxima or minima, we should find out it's optimum value..

Suppose for the maximum value of a..

It's optimum value will be calculated as

$= > \it \: \frac{da}{dt} = 0$
And thus the maximum or minimum value can be calculated by Substituting the value in first equation..
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15)

Given :-
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$= > x = 9 {t}^{2} - {t}^{3}......(1)$
Now,

Given that velocity is maximum.., so

$= > v = \bf \: \frac{dx}{dt} = \frac{d(9 {t}^{2} - {t}^{3} ) }{dt} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: = \it 18t - 3 {t}^{2} .......(2)$

For optimum value,

$= > \frac{dv}{dt} = 0 \\ \\ = > \frac{d(18t - 3 {t}^{2}) }{dt} = 0 \\ \\ = > 18 - 6t = 0 \\ \\ = > \bf \: t = \frac{18}{6} = \underline{3 \: \: sec} \:$

Now putting the value of t in equation (1) we get the position where the velocity is maximum..

So,

$= > \bf x = \it \: 9 {t}^{2} - {t}^{3} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 9 {(3)}^{2} - {(3)}^{3} = 81 - 27 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf \underline{54 \: \: m}$
Hence option a(✔️) is the required position.

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16)

Given equation is for Acceleration..
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So for getting the velocity we have to integrate it..

$= > \bf \frac{dv}{dt} = - k {v}^{3} \\ \\ = > - \frac{dv}{ {v}^{3} } = kdt \\ \\ = > \int( - \frac{1}{ {v}^{3} } )dv = k \int \: dt \\ \\ = > - ( - \frac{1}{2 {v}^{2} } ) = kt \\ \\ = > \frac{1}{ {v}^{2} } = 2kt \\ \\ \bf since \: \: its \: \: limited \: \: so \: \: it \: \: must \: \: have \\ \bf definite \: \: upper \: \: and \: \: lower \: \: values. \\ \\ = > \frac{1}{ {v_{x} } ^{2} } - \frac{1}{ { v_{o} }^{2} } = 2k(t - 0) \\ \\ = > \frac{1}{ { v_{x} }^{2} } = 2kt + \frac{1}{ { v_{o} }^{2} } \\ \\ = > \frac{1}{ { v_{x}}^{2} } = \frac{(2kt + { v_{o}}^{2}) }{ { v_{o} }^{2} } \\ \\ = > { v_{x}}^{2} = \frac{ { v_{o} }^{2} }{2kt + { v_{o} }^{2} } \\ \\ = > \bf v_{x} = \frac{ v_{o}}{ \sqrt{2kt + { v_{o} }^{2} } }$

Hence option c (✔️) is the required velocity.
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17)

Given:
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=> f is the acceleration

=> t is the time

=> T and $f_{o}$ are the constants.
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Since f is Acceleration so integral of f will give the velocity.

So,

$= > f = f_{o}(1 - \frac{t}{ T} ) \\ \\ = > \int \frac{dv}{dt} .dt = f_{o}( \int \: dt - \frac{1}{ T} \int \: tdt) \\ \\ = > \int \: dv = f_{o}(t - \frac{1}{ T} \times \frac{ {t}^{2} }{2} ) \\ \\ = > v = f_{o}t(1 - \frac{t}{ 2T} ) + c$

Thus when t =0, v = 0

So,

$= > 0 = f_{o} \times 0 \times (1 - 0) + c \\ \\ = > c = 0$

Now..

When f =0,... t =?

$= > f = f_{o}(1 - \frac{t}{ T } ) \\ \\ = > \frac{0}{ f_{o}} = 1 - \frac{t}{ T} \\ \\ = > 1 = \frac{t}{ T} \\ \\ = > \bf \: t = \underline{T}$

Now at f = 0.. The particle's velocity will be

$= > \it v_{x} = f_{o}t (1 - \frac{t}{2 T } ) + c \\ \\ = > v_{x} = f_{o}T(1 - \frac{T}{2T} ) + 0 \\ \\ = > \bf v_{x} = f_{o}T( \frac{2 - 1}{2} ) = \frac{1}{2} f_{o} T$

Hence, option c(✔️) is the required velocity.
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