Physics - NLM Question with Solution | TestHub

From$\overrightarrow{F}=\left(10\hat{\mathrm{i}}+5\hat{\mathrm{j}}\right) \text{N}$, we have

$F_x=10\text{ N}\text{; }{F}_y=5\text{ N}$

$a{}_x=\frac{F{}_x}{m}=\frac{10}{\text{0.01}}=1000{\text{ m s}}^{-2}$

As acceleration along the x-axis is constant,

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

$=0+\frac{1}{2}\times 1000\times 5^2=12500\text{ m}$

Similarly,$a{}_y=\frac{F{}_y}{m}=\frac{5}{\text{0.01}}=500{\text{ m s}}^{-2}$

and$y=u_{y}t+\frac{1}{2}{a}_y{t}^2=0+\frac{1}{2}\times 500\times 5^2=6250\text{ m}$

Hence the position of the particle at$t=5 \mathrm{s}$is

$\overrightarrow{r}=\left(12500\hat{\mathrm{i}}+6250\hat{\mathrm{j}}\right) \text{m}$