A particle moves along a straight line in such a way that it’s acceleration is increasing at the rate of 2m / s3 . It’s initial acceleration and velocity were 0, the distance covered by it in t =3 second is,

(B) From given information,dadt=2⇒∫0a da=∫0t 2dt⇒a=2t ⇒dvdt=2t⇒∫0t dv=∫0t 2tdt⇒v=t2⇒drdt=t2∫0r dr=∫03 t2dt⇒r=9m

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A particle moves along a straight line in such a way that it’s acceleration is increasing at the rate of 2m / s³ . It’s initial acceleration and velocity were 0, the distance covered by it in t =3 second is,

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answer is 9.

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Detailed Solution

(B) From given information, $\frac{da}{dt} = 2$
$\Rightarrow \int_{0}^{a} da = \int_{0}^{t} 2 dt \Rightarrow a = 2 t$ $\Rightarrow \frac{dv}{dt} = 2 t \Rightarrow \int_{0}^{t} dv = \int_{0}^{t} 2 tdt$
$\Rightarrow v = t^{2}$ $\Rightarrow \frac{dr}{dt} = t^{2} \int_{0}^{r} dr = \int_{0}^{3} t^{2} dt \Rightarrow r = 9 m$

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