Question

$A,B$ and $C$ are three objects each moving with constant velocity. $A^{\prime }$s speed is $10\text{m/s}$ in a direction $PQ$. The velocity of $B$ relative to $A$ is $6\text{m/s}$ at an angle of ${\cos }^{-1}\left(15/24\right)$ to $PQ$. The velocity of $C$ relative to $B$ is $12\text{m/s}$ in the direction $QP$, then find the magnitude of velocity of $C$.

• A. $3\text{m/s}$
• B. $5\text{m/s}$
• C. $1\text{m/s}$
• D. $7\text{m/s}$

Answer 1

The correct option is B $5\text{m/s}$

$\overrightarrow{V_A}=10\hat{i}$
${\overrightarrow{V}}_{BA}=6\cos \alpha \hat{i}+6\sin \alpha \hat{j}$
Here, $\cos \alpha =\frac{15}{24}$
$\therefore \sin \alpha =\frac{\sqrt{351}}{24}$
and ${\overrightarrow{V}}_{CB}=-12\hat{i}$
$\therefore {\overrightarrow{V}}_{BA}={\overrightarrow{V}}_B-{\overrightarrow{V}}_A$
or $6\cos \alpha \hat{i}+6\sin \alpha \hat{j}=V_B-10\hat{i}$
${\overrightarrow{V}}_B=6\cos \alpha \hat{i}+6\sin \alpha \hat{j}+10\hat{i}$
$=6\times \frac{15}{24}\hat{i}+6\frac{\sqrt{351}}{24}\hat{j}+10\hat{i}$
$=(\frac{15}{4}+10)\hat{i}+\frac{\sqrt{351}}{4}\hat{j}$
$=\frac{55}{4}\hat{i}+\frac{\sqrt{351}}{4}\hat{j}$
But ${\overrightarrow{V}}_{CB}={\overrightarrow{V}}_C-{\overrightarrow{V}}_B$
$\therefore {\overrightarrow{V}}_C={\overrightarrow{V}}_{CB}+{\overrightarrow{V}}_B$
$=-12\hat{i}+\frac{55}{4}\hat{i}+\frac{\sqrt{351}}{4}\hat{j}$
$=\frac{7}{4}\hat{i}+\frac{\sqrt{351}}{4}\hat{j}$
$\therefore V_C=\sqrt{{\left(\frac{7}{4}\right)}^2+{\left(\frac{\sqrt{351}}{4}\right)}^2}$
$=\frac{\sqrt{49+351}}{4}=\frac{\sqrt{400}}{4}=\frac{20}{4}$
$=5\text{m/s}$