A point initially at rest moves along $x$-axis. Its acceleration varies with time as $a - (6t + 5)$in $ms^{- 2}.$ If it starts from origin, the distance covered in 2s is

Question

Dr. Abubacker · Accepted Answer

(b) Given, $a = \frac{dv}{dt} = 6t + 5$
Integrating it $\int_{0}^{v}{dv = \int_{0}^{t}{(6t + 5)dt}}$
$v = \frac{6t^{2}}{2} + 5t$
As $v = \frac{ds}{dt},$ so $ds = \left( \frac{6t^{2}}{2} + 5t ight)dt$
$s = 3\frac{t^{3}}{3} + \frac{5t^{2}}{2}$
Where $t = 2s,s = 3 	imes \frac{2^{3}}{3} + \frac{5 	imes 2^{2}}{2} = 18$m

Q. A point initially at rest moves along $x$-axis. Its acceleration varies with time as $a - (6t + 5)$in $ms^{- 2}.$ If it starts from origin, the distance covered in 2s is

Explanation

Reviewed by Dr. Abubacker

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