Second Law of Motion Questions in English

(A) Given: Mass $m = 1 \ kg$, Force $F = kt$, where $k = 1 \ Ns^{-1}$.
According to Newton's second law, $F = ma$.
Since $a = \frac{dv}{dt}$, we have $m \frac{dv}{dt} = kt$.
Substituting $m = 1$ and $k = 1$, we get $\frac{dv}{dt} = t$.
Integrating with respect to time $t$ (starting from rest, $v=0$ at $t=0$):
$v = \int t \ dt = \frac{t^2}{2}$.
Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \frac{t^2}{2}$.
Integrating again to find the displacement $x$ at $t = 6 \ s$:
$x = \int_{0}^{6} \frac{t^2}{2} \ dt = \left[ \frac{t^3}{6} \right]_{0}^{6}$.
$x = \frac{6^3}{6} = \frac{216}{6} = 36 \ m$.

(A) Given mass $m = 5 \text{ kg}$.
The two forces are $F_1 = 8 \text{ N}$ and $F_2 = 6 \text{ N}$,which are mutually perpendicular.
The resultant force $F$ is given by $F = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}$.
Using Newton's second law,the acceleration $a = F/m = 10 \text{ N} / 5 \text{ kg} = 2 \text{ ms}^{-2}$.
The direction $\theta$ of the resultant force with respect to the $8 \text{ N}$ force is given by $\tan \theta = \frac{F_2}{F_1} = \frac{6}{8} = 3/4$.
Therefore,$\theta = \tan^{-1}(3/4)$ with the $8 \text{ N}$ force.
Thus,the correct option is $A$.