Physics - NLM Question with Solution | TestHub

$F=20 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$

$\therefore$ $F_x=20 \mathrm{N}$ and $F_y=10 \mathrm{N}$

$\therefore$ Acceleration,

$a_x=\frac{F_x}{m}=\frac{20}{1}$
$a_x=20 \mathrm{m} {\mathrm{s}}^{-2}$

and $a_y=\frac{F_y}{m}=\frac{10}{1}=10 \mathrm{m} {\mathrm{s}}^{-2}$

$\therefore$ Displacement of body in $x$ -direction in $2 \mathrm{s}$.

$x=u_x t+\frac{1}{2}{a}_x t^2$

$=0+\frac{1}{2}\times 20\times 2^2$
$=40 \mathrm{m}$

Displacement of body in $y$-direction in $2 \mathrm{s}$.

$y=u_y t+\frac{1}{2} a_y t^2$

$=0+\frac{1}{2}\times 10\times 2^2$

$=20 \mathrm{m}$

$\therefore$ Position of the body at $t=2 \mathrm{s}$

$r=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}$
$r=40 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}$