Answer: 50m along $\tan^{-1} \large\frac{4}{3}$ with x-axis

Acceleration $a=\large\frac{\hat F}{m}=\large\frac{6 \hat i+8 \hat j}{10}$

Since we know the acceleration, we can use the Kinematics equation $ \overrightarrow s=\overrightarrow u t+\large\frac{1}{2} $$ \overrightarrow a t^2$ to get the displacement.

$\quad=0+\large\frac{1}{2}\bigg(\frac{6 \hat i+8 \hat j}{10}\bigg)$$10 ^2 $$=30 \hat i+40 \hat j$

Magnitude $= \sqrt {30 ^2 +40 ^2}= 50\;m$

The direction is along an angle of $\tan ^{-1}\large\frac{40}{30}$$=\tan ^{-1} \;\large\frac{4}{3}$ with x-axis