Question

Two objects A and B are moving along the directions as shown in the figure. Find the magnitude and direction of the relative velocity of B w.r.t. A.

Answer 1

Hint Relative velocity of B w.r.t. A is given by:
 $\Rightarrow \overrightarrow{v_{BA}}=\overrightarrow{v_B}-\overrightarrow{v_A}$
Magnitude of this relative velocity is $\left|\overrightarrow{v_{BA}}\right|$
Direction of this relative velocity is given by the angle $\alpha$ which is calculated by:
 $\Rightarrow \tan \alpha =\frac{v_{B{A}_y}}{v_{B{A}_x}}$
Where $v_{B{A}_y}$ and $v_{B{A}_x}$ are the y and x components of $\overrightarrow{v_{BA}}$ .

Complete step by step solution
 $\begin{array}{rl}& \Rightarrow \overrightarrow{v_A}=10\hat{i}\\ & \Rightarrow \text{Here taking the components of velocity of B;}\\ & \Rightarrow \overrightarrow{v_B}=20\cos 30{}^{\circ }\hat{i}+20\sin 30{}^{\circ }\hat{j}\\ & \Rightarrow 10\sqrt{3}\hat{i}+10\hat{j}\end{array}$
Relative velocity of B w.r.t. A is
 $\begin{array}{rl}& \Rightarrow \overrightarrow{v_{BA}}=\overrightarrow{v_B}-\overrightarrow{v_A}\\ & \Rightarrow =10\sqrt{3}\hat{i}+10\hat{j}-10\sqrt{3}\hat{i}\\ & \Rightarrow =10(\sqrt{3}-1)\hat{i}+10\hat{j}\end{array}$
 $\begin{array}{rl}& Now\\ & \Rightarrow \left|\overrightarrow{v_{BA}}\right|=10\sqrt{{(\sqrt{3}-1)}^2+1^2}\\ & \Rightarrow 10\sqrt{3+1-2\sqrt{3}+1}\\ & \Rightarrow \left|\overrightarrow{v_{BA}}\right|=10\sqrt{5-2\sqrt{3}}m{s}^{-1}\\ & \text{For direction;}\\ & \Rightarrow \tan \alpha =\frac{10}{10(\sqrt{3}-1)}\\ & \Rightarrow \tan \alpha =\frac{1}{\sqrt{3}-1}\\ & \Rightarrow \alpha ={\tan }^{-1}\left(\frac{1}{\sqrt{3}-1}\right)\end{array}$ .

Note
Alternate method:
Velocity of B w.r.t. A:
 $\Rightarrow \overrightarrow{v_{BA}}=\overrightarrow{v_B}+(-\overrightarrow{v_A})$

From the figure;
 $\begin{array}{rl}& \Rightarrow NS=MP=20\sin 30{}^{\circ }\\ & \Rightarrow NS=10\\ & and\\ & \Rightarrow ON=OM-NM\\ & \Rightarrow ON=20\cos 30{}^{\circ }-10\\ & \Rightarrow ON=10(\sqrt{3}-1)\\ & \Rightarrow \left|\overrightarrow{v_{BA}}\right|=\sqrt{O{N}^2+N{S}^2}\\ & \Rightarrow \left|\overrightarrow{v_{BA}}\right|=10\sqrt{{(\sqrt{3}-1)}^2+1^2}\\ & \Rightarrow \left|\overrightarrow{v_{BA}}\right|=10\sqrt{5-2\sqrt{3}}m{s}^{-1}\end{array}$
 $\begin{array}{rl}& \text{For direction;}\\ & \Rightarrow \text{tan }\alpha =\frac{NS}{ON}\\ & \Rightarrow \tan \alpha =\frac{10}{10(\sqrt{3}-1)}\\ & \Rightarrow \tan \alpha =\frac{1}{\sqrt{3}-1}\\ & \Rightarrow \alpha ={\tan }^{-1}\left(\frac{1}{\sqrt{3}-1}\right)\end{array}$ .