Question

explain this question please!

Answer 1

Hope u like my process
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=> The acceleration varies with time (t) as bt.

So,..

$= > \bf \: \frac{d(v)}{dt} = \it \: \blue{bt}$

=> Velocity at time 0 is $\bf \blue{v _{o}}$

Now..

Integrating the change in Acceleration
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We get,

$= > \int \: \it \: dv= \int(bt)dt = \blue{b} \int \blue{t}dt \\ \\ = > \bf \: v = \blue{b \frac{ {t}^{2} }{2} + u} \\ \\ where \: \bf \: v \: \: and \: \: u = \it velocity \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \: \: \: \: \: \: \: \: \underline{ now }\: \\ \\ at \: \: time \: \: (t) = \underline{\blue{0}} \\ \\ = > \bf v_{o} = 0 + u \\ \\ = > u = \underline{ \bf \blue{ v_{o}}}$

Now..
To get the distance trvalled we have to again integrate the velocity.

So..

$= > v = \frac{ds}{dt} = \blue{ \frac{1}{2}b {t}^{2} + v_{o} } \\ \\ = > \int \: ds \: = \blue{ \frac{1}{2}b } \int \: \blue{{t}^{2} } \: dt + \blue{ v_{o}} \int \: dt \\ \\ = > distance(s) = \blue {(\frac{1}{2} \times b \times \frac{ {t}^{3} }{3} )+ v_{o}t } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \blue{ \bold{ v_{o}(t) + \frac{1}{6} b {t}^{3} }}$
At time (t) =0 distance travelled will be 0, so there is no constant term after 2nd integration.
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Hence option c (✔️) is correct.
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Hope this is ur required answer❤️

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