Question

three particles start moving simultaneously from a point on a horizontal smooth plane . first particle moves with speed v₁ toward east , second moves towards north with speed v₂ and third one moves towards north east .what should be the velocity of third particle , so that the three always lie on the same line?

Answer 1

$$\begin{gathered} \mathrm{Coordinates}\mathrm{of}\mathrm{first}\mathrm{particle}\mathrm{after}1\mathrm{second}: \\ \mathrm{First}: \\ \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left({\mathrm{v}}_1,0\right) \\ \mathrm{Second}: \\ \left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(0,{\mathrm{v}}_2\right) \\ \mathrm{Third}: \\ \left({\mathrm{x}}_3,{\mathrm{y}}_3\right)=\left(\frac{{\mathrm{v}}_3}{\sqrt{2}},\frac{{\mathrm{v}}_3}{\sqrt{2}}\right) \\ \mathrm{Condition}\mathrm{for}\mathrm{all}\mathrm{three}\mathrm{particles}\mathrm{to}\mathrm{lie}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{line}: \\ \mathrm{Slopes}\mathrm{must}\mathrm{be}\mathrm{equal}. \\ \frac{{\mathrm{y}}_3-{\mathrm{y}}_1}{{\mathrm{x}}_3-{\mathrm{x}}_1}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1} \\ \frac{\frac{{\mathrm{v}}_3}{\sqrt{2}}-0}{\frac{{\mathrm{v}}_3}{\sqrt{2}}-{\mathrm{v}}_1}=\frac{{\mathrm{v}}_2-0}{0-{\mathrm{v}}_1} \\ \frac{\frac{{\mathrm{v}}_3}{\sqrt{2}}}{\frac{{\mathrm{v}}_3}{\sqrt{2}}-{\mathrm{v}}_1}=-\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1} \\ \frac{\frac{{\mathrm{v}}_3}{\sqrt{2}}}{{\mathrm{v}}_2}=-\frac{\frac{{\mathrm{v}}_3}{\sqrt{2}}-{\mathrm{v}}_1}{{\mathrm{v}}_1} \\ \frac{\frac{{\mathrm{v}}_3}{\sqrt{2}}}{{\mathrm{v}}_2}+\frac{\frac{{\mathrm{v}}_3}{\sqrt{2}}}{{\mathrm{v}}_1}=\frac{{\mathrm{v}}_1}{{\mathrm{v}}_1} \\ \frac{{\mathrm{v}}_3}{\sqrt{2}}\left(\frac{1}{{\mathrm{v}}_2}+\frac{1}{{\mathrm{v}}_1}\right)=1 \\ {\mathrm{v}}_3=\frac{\sqrt{\mathbf{2}}{\mathbf{v}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{2}}}{{\mathbf{v}}_{\mathbf{1}}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}} \end{gathered}$$