1. Tardigrade
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  4. Two particles A and B are located at points (0 , 10 √3) and (.0, 0.) in x-y plane. They start moving simultaneously at the time t=0 s with constant velocities overset arrow VA=5 hati m s- 1 and overset arrow VB=-5√3 hati 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.

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Q. Two particles $A$ and $B$ are located at points $\left(0 , \, 10 \sqrt{3}\right)$ and $\left(\right.0, \, 0\left.\right)$ in $x-y$ plane. They start moving simultaneously at the time $t=0 \, s$ with constant velocities $\overset{ \rightarrow }{V}_{A}=5 \, \hat{i} \, m \, s^{- 1}$ and $\overset{ \rightarrow }{V}_{B}=-5\sqrt{3}\hat{i} \, m \, s^{- 1}$ , 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.

NTA AbhyasNTA Abhyas 2020Motion in a Plane

Solution:

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