Question

Two particles $\text{P}$ and $\text{Q}$ are moving with velocities of $(\hat{i}+\hat{j})$ and $(-\hat{i}+2\hat{j})$ respectively. At time $t=0$, $\text{P}$ is at origin and $\text{Q}$ is at a point with position vector $(2\hat{i}+\hat{j})$. The equation of the path of $\text{Q}$ with respect to $\text{P}$ is $x+2y=C$. Find the value of $C$.

• A. $1$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is C $4$

Given:
Velocity of point $\text{P},$
$\overrightarrow{v_P}=(\hat{i}+\hat{j})$
Velocity of point $\text{Q},$
$\overrightarrow{v_Q}=(-\hat{i}+2\hat{j})$


At the time $t=0$
$\overrightarrow{r_P}=0;\overrightarrow{r_Q}=(2\hat{i}+\hat{j})$

So, at any time $t$, position of point $\text{P \& Q}$

$\overrightarrow{r_p}=(\hat{i}+\hat{j})t$

$\overrightarrow{r_Q}=(2\hat{i}+\hat{j})+(-\hat{i}+2\hat{j})t=(2-t)\hat{i}+(1+2t)\hat{j}$

So, position of particle $\text{Q}$ w.r.t. $\text{P}$

${\overrightarrow{r}}_{QP}=\overrightarrow{r_Q}-\overrightarrow{r_P}$

$\Rightarrow {\overrightarrow{r}}_{QP}=(2-t)\hat{i}+(1+2t)\hat{j}-(\hat{i}+\hat{j})t$

$\Rightarrow {\overrightarrow{r}}_{QP}=(2-2t)\hat{i}+(1+t)\hat{j}......\left(1\right)$

From eq. $\left(1\right)$,

$x=2-2t;y=1+t$

$\therefore x+2y=(2-2t)+2(1+t)=4$

And given equation is $x+2y=C$

Therefore, $C=4$


$$\begin{array}{c}\text{Why this question?}\\ \text{Relative position,}\overrightarrow{r_{AB}}=\overrightarrow{r_A}-\overrightarrow{r_B}\\ \text{Relative velocity,}\overrightarrow{v_{AB}}=\overrightarrow{v_A}-\overrightarrow{v_B}\end{array}$$