Thank you for reporting, we will resolve it shortly
Q.
A block of mass $1 \,kg$ starts from rest at $x\, = \,0$ and moves along the x-axis under the action of a force $F = kt$, where $t$ is time and $k\, = \,1 \,Ns^{-1}$. The distance the block will travel in $6$ seconds is
From Newton's second law
$F=m a $
$m=1\, kg $
$F=1 \times a=\frac{d v}{d t}$
According to the question,
$F=k t=\frac{d v}{d t}$
$d v=k t \cdot d t$
Integration gives
$v =\frac{k t^{2}}{2}=\frac{d x}{d t} $
$ x =\frac{k t^{3}}{6}=\frac{1 \times 6 \times 6 \times 6}{6} $
$=36\, m $