Q124 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Physics
Two cars $A$ and $B$ are moving with speeds $v_A=120 \mathrm{~km} / \mathrm{h}$ and $v_B=50 \mathrm{~km} / \mathrm{h}$ respectively in the directions as indicated by the arrow in the figure below. What is the relative speed of the car $B$ with respect to car $A$ ?
- A$70 \mathrm{~km} / \mathrm{h}$
- B$120 \mathrm{~km} / \mathrm{h}$
- Correct answerC$130 \mathrm{~km} / \mathrm{h}$
- D$170 \mathrm{~km} / \mathrm{h}$
Solution
The given situation is shown in the following figure
Speed of car $A$, $v_A=120 \mathrm{~km} / \mathrm{h} \Rightarrow \mathbf{v}_A=-120 \hat{\mathbf{i}}$ Speed of car $B$, $v_B=50 \mathrm{~km} / \mathrm{h}$ $\mathbf{v}_B=-50 \hat{\mathbf{j}}$ $\therefore$ Relative speed of the car $B$ w.r.t. car $A$, $\mathbf{v}_{B A}=\mathbf{v}_B-\mathbf{v}_A=-50 \hat{\mathbf{j}}-(-120 \hat{\mathbf{i}})$ $=-50 \hat{\mathbf{j}}+120 \hat{\mathbf{i}}=120 \hat{\mathbf{i}}-50 \hat{\mathbf{j}}$ $\therefore \quad\left|\mathbf{v}_{B A}\right|=\sqrt{(120)^2+(-50)^2}$ $=\sqrt{16900}=130 \mathrm{~km} / \mathrm{h}$
Speed of car $A$, $v_A=120 \mathrm{~km} / \mathrm{h} \Rightarrow \mathbf{v}_A=-120 \hat{\mathbf{i}}$ Speed of car $B$, $v_B=50 \mathrm{~km} / \mathrm{h}$ $\mathbf{v}_B=-50 \hat{\mathbf{j}}$ $\therefore$ Relative speed of the car $B$ w.r.t. car $A$, $\mathbf{v}_{B A}=\mathbf{v}_B-\mathbf{v}_A=-50 \hat{\mathbf{j}}-(-120 \hat{\mathbf{i}})$ $=-50 \hat{\mathbf{j}}+120 \hat{\mathbf{i}}=120 \hat{\mathbf{i}}-50 \hat{\mathbf{j}}$ $\therefore \quad\left|\mathbf{v}_{B A}\right|=\sqrt{(120)^2+(-50)^2}$ $=\sqrt{16900}=130 \mathrm{~km} / \mathrm{h}$
Hence, the correct answer is option: 'C'.