Question

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

Answer 1

Analytically:

${\overrightarrow{v}}_A=10\hat{i},{\overrightarrow{v}}_B=20\cos {30}^o\hat{i}+20\sin {30}^o\hat{j}=10\sqrt{3}\hat{i}+10\hat{j}$

Relative velocity of $B$ w.r.t. $A$

${\overrightarrow{v}}_{B/A}={\overrightarrow{v}}_B-{\overrightarrow{v}}_A=10(\sqrt{3}-1)\hat{i}+10\hat{j}$

$\Rightarrow |{\overrightarrow{v}}_{B/A}|=10\sqrt{(\sqrt{3}-1{)}^2+1^2}=10\sqrt{5-2\sqrt{3}}m{s}^{-1}$

$\tan \alpha =\frac{10}{10(\sqrt{3}-1)}=\frac{1}{\sqrt{3}-1}$

Graphically: Velocity of $B$ with respect to $A$

${\overrightarrow{v}}_{B/A}={\overrightarrow{v}}_B+(-{\overrightarrow{v}}_A)$

From Fig., $NS=MP=20\sin {30}^o=10$

and $ON=OM-NM=20\cos {30}^o-10=10(\sqrt{3}-1)$

$\Rightarrow |{\overrightarrow{v}}_{B/A}|=\sqrt{O{N}^2+N{S}^2}$

$=10\sqrt{(\sqrt{3}-1{)}^2+(1{)}^2}$

$=10\sqrt{5-2\sqrt{3}}m{s}^{-1}$

$tan\alpha =\frac{NS}{ON}=\frac{10}{10(\sqrt{3}-1)}=\frac{1}{\sqrt{3}-1}$