A force $\overrightarrow{\mathrm{F}}=(40 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}) \mathrm{N}$ acts on a body of mass
$5 \mathrm{~kg}$. If the body starts from rest, its position vector $\overrightarrow{\mathrm{r}}$ at time $\mathrm{t}=10 \mathrm{~s}$, will be :
Correct Option: , 3
$\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=(8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}$
$\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=\overrightarrow{\mathrm{v}}=(8 \mathrm{t} \hat{\mathrm{i}}+2 \mathrm{t} \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}$
$\overrightarrow{\mathrm{r}}=(8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \frac{\mathrm{t}^{2}}{2} \mathrm{~m}$
At $\mathrm{t}=10 \mathrm{sec}$
$\overrightarrow{\mathrm{r}}=[(8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}) 50] \mathrm{m}$
$\Rightarrow \vec{r}=(400 \hat{i}+100 \hat{j}) m$
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