Question

Two cars are $15m$ apart as shown in the figure at time $t=0$. Find the closest distance between the cars. (Velocity of cars are as shown in the figure)

• A. $9m$
• B. $12m$
• C. $13m$
• D. $15m$

Answer 1

The correct option is B $12m$



w.r.t. A, motion of $\stackrel{B}{\to }$ is as shown by $V_{BA}$ Shortest distance is AC
$AC=ABcos\theta$
$=15cos\theta$



$\overrightarrow{V_{BA}}=\overrightarrow{V_{BG}}-\overrightarrow{V_{AG}}$
$=3\hat{j}-(-4)\hat{i}$
$\overrightarrow{V_{BA}}=4\hat{i}+3\hat{j}$

$\downarrow \downarrow$
${\left(V_{BA}\right)}_x{\left(V_{BA}\right)}_y$

$\tan \theta =\frac{{\left(V_{BA}\right)}_y}{{\left(V_{BA}\right)}_x}=\frac{3}{4}$
$\Rightarrow \theta ={37}^{\circ }$
$\therefore AC=15\times \frac{4}{5}$
$AC=12m$