Car $A$ is moving to the East with a speed of $30 \mathrm{~km} / \mathrm{hr}$, and car $B$ is moving to the North with the same speed. What is the velocity of car $B$ as measured in Car $A$ ?

• Correct answer
  A

  $42 \mathrm{~km} / \mathrm{hr}, 45$ North of West
• B

  $42 \mathrm{~km} / \mathrm{hr}, 45$ East of North
• C

  $60 \mathrm{~km} / \mathrm{hr}, 45$ South of East
• D

  $42 \mathrm{~km} / \mathrm{hr}, 45$ South of East

Solution

Given that, velocity of $\operatorname{car} A$, $$ \begin{aligned} & v_A=30 \mathrm{~km} / \mathrm{h} \text { (toward east) } \\ & \Rightarrow \mathbf{v}_A=30 \hat{\mathbf{i}} \end{aligned} $$ For car $B, v_B=30 \mathrm{~km} / \mathrm{h}$ (toward North) $$ \Rightarrow \quad \mathbf{v}_B=30 \hat{\mathbf{j}} $$
According to question, Velocity of car $B$ as measured in car $A=$ relative velocity of car $B$ w.r.t car $A$. $$ \mathbf{v}_{B A}=\mathbf{v}_B-\mathbf{v}_A=30 \hat{\mathbf{j}}-30 \hat{\mathbf{i}} $$ $\therefore$ Magnitude of $\mathbf{v}_{B A}$ $$ \begin{aligned} & =\sqrt{v_B^2+v_A^2}=\sqrt{30^2+30^2} \\ & =30 \sqrt{2} \mathrm{~km} / \mathrm{h} \approx 42 \mathrm{~km} / \mathrm{h} \end{aligned} $$ Direction of $\mathbf{v}_{B A}$, $$ \begin{aligned} \tan \theta & =\frac{v_B}{v_A}=\frac{30}{-30}=-1 \\ \Rightarrow \quad \theta & =135^{\circ} \end{aligned} $$ Hence, velocity of car $B$ as measured in car $A$ is $42 \mathrm{~km} / \mathrm{h}$ at angle of $\left(135^{\circ}-90^{\circ}=45^{\circ}\right)$ North of West.

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