Question

A particle moves from the point $(2.0\hat{i}+4.0\hat{j})\text{m}$, at $t=0$, with an initial velocity $(5.0\hat{i}+4.0\hat{j}){\text{ms}}^{-1}$. It is acted upon by a constant force, which produces a constant acceleration of $(4.0\hat{i}+4.0\hat{j}){\text{ms}}^{-2}$. What is the distance of the particle, from the origin, at time $2s$?

• A. $5\text{m}$
• B. $10\sqrt{2}\text{m}$
• C. $20\sqrt{2}\text{m}$
• D. $15\text{m}$

Answer 1

The correct option is C $20\sqrt{2}\text{m}$

As $\overrightarrow{S}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$
$\overrightarrow{S}=(5\hat{i}+4\hat{j})2+\frac{1}{2}(4\hat{i}+4\hat{j})4$
$=10\hat{i}+8\hat{j}+8\hat{i}+8\hat{j}$
$\therefore \overrightarrow{S}={\overrightarrow{r}}_f-{\overrightarrow{r}}_i=18\hat{i}+16\hat{j}$
Here, ${\overrightarrow{r}}_i=(2.0\hat{i}+4.0\hat{j})$
$\Rightarrow {\overrightarrow{r}}_f=(20\hat{i}+20\hat{j})$
so that, $(r_f{)}_x=20,(r_f{)}_y=20$

Now, distance of the particle from the origin is,
$d=\sqrt{(r_f{)}_x^2+(r_f{)}_y^2}=\sqrt{(20{)}^2+(20{)}^2}$

$=\sqrt{400+400}=20\sqrt{2}\text{m}$

Hence, $\left(B\right)$ is the correct answer.