Physics - Kinematics Question with Solution | TestHub

Let the x -axis be eastwards and the y -axis be northwards ${\vec{v}}_1=30\hat{i}$ and ${\vec{v}}_2=40\hat{j}$ ${\vec{v}}_{21}$ (relative velocity of 2 in relation to 1)is given by ${\vec{v}}_2-{\vec{v}}_1=40\hat{j}-30\hat{i}$

From the concept relative velocity, ship 1 may be assumed to be fixed and ship 2 to be moving with the velocity $(-30\hat{i}+40\hat{j})$. This is a vector of magnitude 50 and inclined at $\theta ={\tan }^{-1}(4/3)$. The shortest distance is obviously the perpendicular distance of the relative velocity vector from the fixed position of ship 1 . Therefore, $s_{\text{shortest }}=s_{12}\sin \theta$ where $s_{12}$ is the initial distance between 1 and 2 is $10\times \frac{4}{5}=8\mathrm{km}$ \end{document}