Physics - Kinematics Question with Solution | TestHub

To find the relative velocity ${\vec{v}}_{A/B}={\vec{v}}_A-{\vec{v}}_B$.

${\vec{v}}_A=80\cos 45^{\circ }\hat{j}+80\sin 45^{\circ }\hat{i}=40\sqrt{2}\hat{j}+40\sqrt{2}\hat{i}$

${\vec{v}}_B=60\cos 45^{\circ }(-\hat{j})+60\sin 45^{\circ }\hat{i}=-30\sqrt{2}\hat{j}+30\sqrt{2}\hat{i}$

${\vec{v}}_{A/B}=(40\sqrt{2}-30\sqrt{2})\hat{i}+(40\sqrt{2}-(-30\sqrt{2}))\hat{j}=10\sqrt{2}\hat{i}+70\sqrt{2}\hat{j}$

$\tan \alpha =\frac{({\vec{v}}_{A/B}{)}_x}{({\vec{v}}_{A/B}{)}_y}=\frac{10\sqrt{2}}{70\sqrt{2}}=\frac{1}{7}$