Question bank/JEE Main/Physics

Initially two particles A and B are present at (0,0) and (d, 0)
respectively. They start moving with speed
${\overset{\rightarrow}{v}}_{A} = v\overset{\hat{}}{i} + v\overset{\hat{}}{j}$and$V_{B} = - v\overset{\hat{}}{j}$.
If R is the separation between them and $T_{0}$ be the time when
separationbetween them is minimum, then

A$T_{0} = \frac{d}{5v}$B$R_{\min} = \frac{d}{\sqrt{5}}$CGraph of R vs time is straight lineDGraph of R vs time is Circle

Show answer and solution▾

$V_{BA} = - v\overset{\hat{}}{i} - 2v\overset{\hat{}}{j}$
$tan\theta = 2$
[IMAGE] at any time t
$x = d - vt$
$y = - 2vt$
$\therefore R^{2} = (d - vt)^{2} + 4v^{2}t^{2}$
$R = \sqrt{5v^{2}t^{2} - 2vdt + d^{2}}$
It is neither st. line nor circle
$R_{\min} = dsin\theta = \frac{2d}{\sqrt{5}}$
Distance $BC = dcos\theta$
$= \frac{d}{\sqrt{5}}$
time, $T_{0} = \frac{d}{\sqrt{5}v_{BA}}$
$T_{0} = \frac{d}{\sqrt{5} \times \sqrt{5v}};\ T_{0} = \frac{d}{5v}$

Source: Grass-G / P-05 Physics paper 10 October.docx