Question

If a car is moving with velocity 40km/hr towards East and a person in the car sees a train moving towards north with velocity 40 root 3 km/hr. Find the actual direction and velocity of train.

Answer 1

$$\begin{gathered} \mathrm{Velocity}\mathrm{of}\mathrm{train}\mathrm{wrt}\mathrm{passenger}\mathrm{in}\mathrm{car}: \\ {\overrightarrow{\mathrm{V}}}_{\mathrm{tc}}={\overrightarrow{\mathrm{V}}}_{\mathrm{t}}-{\overrightarrow{\mathrm{V}}}_{\mathrm{c}} \\ {\overrightarrow{\mathrm{V}}}_{\mathrm{t}}={\overrightarrow{\mathrm{V}}}_{\mathrm{tc}}+{\overrightarrow{\mathrm{V}}}_{\mathrm{c}} \\ \left|{\overrightarrow{\mathrm{V}}}_{\mathrm{t}}\right|=\sqrt{{{\mathrm{V}}_{\mathrm{tc}}}^2+{{\mathrm{V}}_{\mathrm{c}}}^2+2{\mathrm{V}}_{\mathrm{tc}}{\mathrm{V}}_{\mathrm{c}}\mathrm{cos\theta }} \\ =\sqrt{{\left(40\sqrt{3}\right)}^2+{\left(40\right)}^2+2{\mathrm{V}}_{\mathrm{tc}}{\mathrm{V}}_{\mathrm{c}}\cos 90} \\ =80\mathrm{km}/\mathrm{hr} \\ \mathrm{Direction}\mathrm{of}\mathrm{train}:(\mathrm{Angle}\mathrm{made}\mathrm{with}\mathrm{direction}\mathrm{of}\mathrm{car}) \\ \mathrm{\varphi }={\tan }^{-1}\frac{{\mathrm{V}}_{\mathrm{tc}}}{{\mathrm{V}}_{\mathrm{c}}} \\ ={\tan }^{-1}\frac{40\sqrt{3}}{40} \\ ={60}^{\mathrm{o}} \end{gathered}$$