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(B) Given,acceleration $a = 4t$.Since the particle starts from rest,the initial velocity $u = 0$ at $t = 0$.We know that $a = \frac{dv}{dt}$,so $dv =

Vedclass · Accepted Answer

$18$

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(B) Given,acceleration $a = 4t$.Since the particle starts from rest,the initial velocity $u = 0$ at $t = 0$.We know that $a = \frac{dv}{dt}$,so $dv = a \ dt = 4t \ dt$.Integrating both sides,we get $v = \int 4t \ dt = 2t^2 + C$.At $t = 0$,$v = 0$,so $C = 0$. Thus,$v = 2t^2$.We know that $v = \frac{dx}{dt}$,so $dx = v \ dt = 2t^2 \ dt$.Integrating both sides to find the distance $d$ covered from the origin ($x=0$ at $t=0$):$d = \int_{0}^{3} 2t^2 \ dt = \left[ \frac{2t^3}{3} ight]_{0}^{3}$.$d = \frac{2(3)^3}{3} = \frac{2 	imes 27}{3} = 2 	imes 9 = 18 \ m$.

$A$ particle initially at rest moves along the $x$-axis. Its acceleration varies with time as $a = 4t$. If it starts from the origin,the distance covered by it in $3 \ s$ is $........... \ m$.

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