Question

Four particles $A,B,C$ and $D$ are in motion. The velocities of one with respect to other are given as ${\overrightarrow{v}}_{DC}$ is $20\text{m/s}$ towards north, ${\overrightarrow{v}}_{BC}$ is $20\text{m/s}$ towards east and ${\overrightarrow{v}}_{BA}$ is $20\text{m/s}$ towards south. Then ${\overrightarrow{v}}_{DA}$ is

• A. $20\text{m/s}$ towards north
• B. $20\text{m/s}$ towards south
• C. $20\text{m/s}$ towards east
• D. $20\text{m/s}$ towards west

Answer 1

The correct option is D $20\text{m/s}$ towards west

From the question, we know that, $\begin{array}{}{\overrightarrow{v}}_{DC}={\overrightarrow{v}}_D-{\overrightarrow{v}}_C=20\hat{j}& \text{(1)}\end{array}$
$\begin{array}{}{\overrightarrow{v}}_{BC}={\overrightarrow{v}}_B-{\overrightarrow{v}}_C=20\hat{i}& \text{(2)}\end{array}$
$\begin{array}{}{\overrightarrow{v}}_{BA}={\overrightarrow{v}}_B-{\overrightarrow{v}}_A=-20\hat{j}& \text{(3)}\end{array}$
Equation $\left(1\right)-\left(2\right)+\left(3\right)$ gives,
${\overrightarrow{v}}_D-{\overrightarrow{v}}_A=-20\hat{i}$
or ${\overrightarrow{v}}_{DA}=-20\hat{i}$
Therefore ${\overrightarrow{v}}_{DA}$ is $20\text{m/s}$ towards west.