Question

Two particles A and B move with velocities $v_{1}and{v}_2$ respectively along the x and y axis. The initial separation between them is 'd' as shown in the figure. Find the least distance between them during their motion

• A. $\frac{d{V}_1^2}{V_1^2+V_2^2}$
• B. $\frac{d{V}_2^2}{V_1^2+V_2^2}$
• C. $\frac{d{V}_2}{\sqrt{V_1^2+V_2^2}}$
• D. $\frac{d{V}_1}{\sqrt{V_1^2+V_2^2}}$

Answer 1

The correct option is C $\frac{d{V}_2}{\sqrt{V_1^2+V_2^2}}$

Let
$\overrightarrow{v_1}=v_1\hat{i}$
$\overrightarrow{v_2}=v_2\hat{j}$
$\overrightarrow{v_2}-\overrightarrow{v_1}=v_2\hat{i}-v_1\hat{j}$

$\overrightarrow{s_2}\left(t\right)-\overrightarrow{s_1}\left(t\right)=d\hat{j}+v_{2}t\hat{i}-v_{1}t\hat{j}$

$|\overrightarrow{s_2}\left(t\right)-\overrightarrow{s_1}\left(t\right){|}^2=(v_{2}t{)}^2+(d-v_{1}t{)}^2$

$=(v_2^2+v_1^2)t^2-2d{v}_{1}t+d^2$

$=(\sqrt{v_1^2+v_2^2}t-\frac{d{v}_1}{\sqrt{v_2^2+v_1^2}}{)}^2+d^2-\frac{d^2{v}_1^2}{v_1^2+v_2^2}$

$\ge \frac{d^2{v}_2^2}{v_1^2+v_2^2}$

$|\overrightarrow{s_1}\left(t\right)-\overrightarrow{s_2}\left(t\right)|\ge \frac{d{v}_2}{\sqrt{v_1^2+v_2^2}}$