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A particle moves from the point $(2.0 \hat{i}+4.0 \hat{j}) \mathrm{m},$ at $\mathrm{t}=0$,
with an initial velocity $(5.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-1} .$ It is acted upon by a constant force which produces a constant acceleration
$(4.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-2} .$ What is the distance of the particle
from the origin at time $2 \mathrm{~s} ?$
Options:
with an initial velocity $(5.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-1} .$ It is acted upon by a constant force which produces a constant acceleration
$(4.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-2} .$ What is the distance of the particle
from the origin at time $2 \mathrm{~s} ?$
Solution:
1919 Upvotes
Verified Answer
The correct answer is:
$20 \sqrt{2} \mathrm{~m}$
As $\overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{ut}}+\frac{1}{2} \overrightarrow{\mathrm{a}} \mathrm{t}^{2}$
$\overrightarrow{\mathrm{S}}=(5 \hat{i}+4 \hat{j}) 2+\frac{1}{2}(4 \hat{i}+4 \hat{j}) 4$
$=10 \hat{i}+8 \hat{j}+8 \hat{i}+8 \hat{j}$
$\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}=18 \hat{\mathrm{i}}+16 \hat{\mathrm{j}}$
$\left[\mathrm{as} \overrightarrow{\mathrm{s}}=\right.$ change in position $\left.=\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}\right]$
$\overrightarrow{\mathrm{r}}_{\mathrm{r}}=20 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}$
$\left|\overrightarrow{\mathrm{r}}_{\mathrm{r}}\right|=20 \sqrt{2}$
$\overrightarrow{\mathrm{S}}=(5 \hat{i}+4 \hat{j}) 2+\frac{1}{2}(4 \hat{i}+4 \hat{j}) 4$
$=10 \hat{i}+8 \hat{j}+8 \hat{i}+8 \hat{j}$
$\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}=18 \hat{\mathrm{i}}+16 \hat{\mathrm{j}}$
$\left[\mathrm{as} \overrightarrow{\mathrm{s}}=\right.$ change in position $\left.=\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}\right]$
$\overrightarrow{\mathrm{r}}_{\mathrm{r}}=20 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}$
$\left|\overrightarrow{\mathrm{r}}_{\mathrm{r}}\right|=20 \sqrt{2}$
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