Two cars, at a certain instant, are $50 \mathrm{~km}$ apart on a line running from south to north. The one farther north is moving west at $25 \mathrm{~km} / \mathrm{hr}$. The other is moving towards north at $25 \mathrm{~km} / \mathrm{hr}$. How long do they take to reach their distance of closest approach?

• A

  30 min
• Correct answer
  B

  60 min
• C

  85 min
• D

  90 min

Solution

Let us sit on B and observe 'A'. Then, $\overrightarrow{\mathrm{V}}_{\mathrm{AB}}=\overrightarrow{\mathrm{V}}_{\mathrm{A}}-\overrightarrow{\mathrm{V}}_{\mathrm{B}}$ $$ \begin{aligned} & \left|\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\right|=\sqrt{25^2+25^2} \\ & =25 \sqrt{2} \mathrm{~km} / \mathrm{hr} \end{aligned} $$ From diagram, we can clearly see when A reaches point ' $\mathrm{C}$ ' its distance is closest from $\mathrm{B}$. $$ \tan \theta=\frac{\left|\mathrm{V}_{\mathrm{B}}\right|}{\left|\mathrm{V}_{\mathrm{A}}\right|}=1 \Rightarrow \theta=45^{\circ} $$ So, $\mathrm{AC}=50 \cos 45^{\circ}=25 \sqrt{2} \mathrm{~km}$ Then, $\mathrm{t}=\frac{\mathrm{AC}}{\left|\mathrm{V}_{\mathrm{AB}}\right|}=\frac{25 \sqrt{2}}{25 \sqrt{2}}=1 \mathrm{hr}=60 \mathrm{~min}$

Hence, the correct answer is option: 'B'.