Question

3 particles start moving simultaneously from a point on a horizontal smooth plane. First particle moves with speed v₁ towards east.,second particle moves with speed v₂ towards north and third one moves north east .Find the velocity of the third particle so that the 3 always lie on the same straight line.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}\mathrm{us}\mathrm{consider}\mathrm{particles}\mathrm{starts}\mathrm{from}\mathrm{origin},\mathrm{positive}\mathrm{x}\mathrm{and}\mathrm{positive}\mathrm{y}\mathrm{are}\mathrm{taken}\mathrm{east}\mathrm{and}\mathrm{north}\mathrm{respectively}. \\ \mathrm{The}\mathrm{position}\mathrm{of}\mathrm{first}\mathrm{particle}\mathrm{at}\mathrm{time}\mathrm{t}\mathrm{is}\left({\mathrm{v}}_1\mathrm{t},0\right) \\ \mathrm{The}\mathrm{position}\mathrm{of}\mathrm{second}\mathrm{particle}\mathrm{at}\mathrm{time}\mathrm{t}\mathrm{is}\left(0,{\mathrm{v}}_2\mathrm{t}\right) \\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{joining}\mathrm{these}\mathrm{two}\mathrm{positions} \\ \mathrm{y}=\left(\frac{{\mathrm{v}}_2\mathrm{t}-0}{0-{\mathrm{v}}_1\mathrm{t}}\right)\left(\mathrm{x}-{\mathrm{v}}_1\mathrm{t}\right)+0 \\ \mathrm{y}=-\left(\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}\right)\left(\mathrm{x}-{\mathrm{v}}_1\mathrm{t}\right) \\ \mathrm{y}=-\left(\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}\right)\mathrm{x}+{\mathrm{v}}_2\mathrm{t}.......\left(1\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{path}\mathrm{of}\mathrm{third}\mathrm{particle} \\ \mathrm{y}=\mathrm{x}....\left(2\right) \\ \mathrm{from}\left(1\right)\&\left(2\right) \\ \mathrm{x}=\mathrm{y}=\left(\frac{{\mathrm{v}}_1{\mathrm{v}}_2}{{\mathrm{v}}_1+{\mathrm{v}}_2}\right)\mathrm{t} \\ \mathrm{distance}\mathrm{of}\mathrm{this}\mathrm{point}\mathrm{from}\mathrm{origin}\mathrm{d}=\sqrt{2}\left(\frac{{\mathrm{v}}_1{\mathrm{v}}_2}{{\mathrm{v}}_1+{\mathrm{v}}_2}\right)\mathrm{t} \\ \mathrm{If}\mathrm{third}\mathrm{particle}\mathrm{attains}\mathrm{such}\mathrm{distance}\mathrm{in}\mathrm{time}\mathrm{t}\mathrm{then}\mathrm{the}\mathrm{will}\mathrm{be}\mathrm{lie}\mathrm{on}\mathrm{a}\mathrm{line},\mathrm{therefore} \\ {\mathrm{v}}_3\mathrm{t}=\mathrm{d} \\ {\mathrm{v}}_3=\frac{\mathrm{d}}{\mathrm{t}} \\ {\mathrm{v}}_3=\sqrt{2}\left(\frac{{\mathrm{v}}_1{\mathrm{v}}_2}{{\mathrm{v}}_1+{\mathrm{v}}_2}\right) \end{gathered}$$