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 acceleration $(4.0\hat{i}+4.0\hat{j}){\text{ms}}^{-2}$. What is the distance of the particle from the origin at time $2\text{s}$?

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

Answer 1

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

Given,

Initial position vector, ${\overrightarrow{r}}_i=(2.0\hat{i}+4.0\hat{j})\text{m}$

Initial velocity, $\overrightarrow{u}=(5.0\hat{i}+4.0\hat{j}){\text{ms}}^{-1}$

Acceleration, $\overrightarrow{a}=(4.0\hat{i}+4.0\hat{j}){\text{ms}}^{-2}$

Applying the equation of motion,

$\overrightarrow{S}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$

Where, $\overrightarrow{S}=\overrightarrow{r_f}-\overrightarrow{r_i}$

Substituting the values,

$\overrightarrow{S}=(5\hat{i}+4\hat{j})2+\frac{1}{2}(4\hat{i}+4\hat{j})2^2$

$=10\hat{i}+8\hat{j}+8\hat{i}+8\hat{j}$

$\Rightarrow {\overrightarrow{r}}_f-{\overrightarrow{r}}_i=18\hat{i}+16\hat{j}$

$\Rightarrow {\overrightarrow{r}}_f-(2\hat{i}+4\hat{j})=18\hat{i}+16\hat{j}$

$\Rightarrow {\overrightarrow{r}}_f=20\hat{i}+20\hat{j}$

$\therefore |{\overrightarrow{r}}_f|=20\sqrt{2}\text{m}$

Hence, $\text{(A)}$ is the correct answer.