Question

A person standing on the ground observe that rain is falling with a speed 10m/s making sin − 1(0.6) with vertical toward west .Now he starts to run towards east with an acceleration of 0.5m/s². Then the distance he had covered by the time he observed the rain to be falling at 45° with the vertical. Find the distance covered by the person?

Answer 1

$\vec{v_R}=-10(\sin(\sin^{-1}(0.6))\hat{i}+\cos(\sin^{-1}(0.6))\hat{j})\\ =-10(0.6\hat{i}+0.8{j})\\ =(-6\hat{i}-8\hat{j})\textrm{m/s}\\ \vec{v_m}=\sqrt{2as}\hat{i}=\sqrt{s}\hat{i}\\ \vec{v_{rm}}=(-6-\sqrt{s})\hat{i}-8\hat{j}\\ \tan 45^{\circ}=\dfrac{8}{6+\sqrt{s}}\\ \Rightarrow \sqrt{s}=2\\ \Rightarrow s=4 \textrm{m}$