# A force

Question:

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 :

1. $(100 \hat{\mathrm{i}}+400 \hat{\mathrm{j}}) \mathrm{m}$

2. $(100 \hat{i}+100 \hat{j}) m$

3. $(400 \hat{\mathrm{i}}+100 \hat{\mathrm{j}}) \mathrm{m}$

4. $(400 \hat{\mathrm{i}}+400 \hat{\mathrm{j}}) \mathrm{m}$

Correct Option: , 3

Solution:

$\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=(8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}$

$\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=\overrightarrow{\mathrm{v}}=(8 \mathrm{t} \hat{\mathrm{i}}+2 \mathrm{t} \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}$

$\overrightarrow{\mathrm{r}}=(8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \frac{\mathrm{t}^{2}}{2} \mathrm{~m}$

At $\mathrm{t}=10 \mathrm{sec}$

$\overrightarrow{\mathrm{r}}=[(8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}) 50] \mathrm{m}$

$\Rightarrow \vec{r}=(400 \hat{i}+100 \hat{j}) m$