Question

Two particles $A$ and $B$ are located at points $\left(0,10\sqrt{3}\right)$ and $(0,0)$ in $x-y$ plane. They start moving simultaneously at the time $t=0s$ with constant velocities ${\overrightarrow{V}}_A=5\hat{i}m{s}^{-1}$ and ${\overrightarrow{V}}_B=-5\sqrt{3}\hat{i}m{s}^{-1}$ , respectively. The time when they are closest to each other is found to be $K/2$ second. Find $K$ . All distances are given in meter.

Answer 1

Answer: 0

As observed by B motion of A is along AM and BM is the shortest distance between them. Relative displacement of A w.r.t to B is $AM=ABcos30^o$

${\overrightarrow{V}}_{AB}=5\hat{i}+5\sqrt{3}\hat{j}$
Time taken $(t)=\frac{\left|{\overrightarrow{S}}_{AB}\right|}{\left|{\overrightarrow{V}}_{AB}\right|}=\frac{AM}{V_{AB}}=\frac{10\sqrt{3}\cos 30^{\circ }}{10}$
$=1.5s$