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Two particles A and B moving in x-y plane are at origin at t = 0 sec. The initial velocity vectors of A and B are $\vec{u_{A}} = 8 \hat{i}$ m/s and $\vec{u_{B}} = 8 \hat{j}$ m/s. The acceleration of A and B are constant and are $\vec{a_{A}} = - 2 \hat{i}$ $m / s^{2}$ and $\vec{a_{B}} = - 2 \hat{j}$ $m / s^{2}$ . Column I gives certain statements regarding particle A and B. Column-II gives corresponding results. Match the statements in column I with corresponding results in Column II.

Column-I
Column-II
A) The time (in seconds) at which velocity of A relative to B is zero p) $16 \sqrt{2}$
B) The distance (in metres) between A and B when their relative velocity is zero. q) $8 \sqrt{2}$
C) The time (in seconds) after t = 0 sec, at which A and B are at same position r) 8
D) The magnitude of relative velocity of A and B at the instant they are at same position. s) 4

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A-s; B-p; C-r; D-Q;

A-p; B-p; C-r; D-Q;

A-s; B-q; C-r; D-Q;

A-s; B-p; C-p; D-Q;

answer is C.

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Detailed Solution

The initial velocity of A relative to B is ${\vec{u}}_{A B} = {\vec{u}}_{A} - {\vec{u}}_{B} = (8 \hat{i} - 8 \hat{j}) m / s$ $u A B = 8 \sqrt{2} m / s$
acceleration of A relative to B is -
${\vec{a}}_{A B} = {\vec{a}}_{A} - {\vec{a}}_{B} (- 2 i + 2 j) m / s^{2} ∴ a_{A B} = 2 \sqrt{2} m / s^{2}$
since B observes initial velocity and constant acceleration of A in opposite directions, Hence B observes A moving along a straight line.
From frame of B
Hence time when vAB = 0 is t = $\frac{u_{A B}}{a_{A B}} =$ 4 sec.
The distance between A & B when vAB = 0 is

S = $\frac{{u^{2}}_{A B}}{2 a_{A B}} =$ m
The $\frac{2 u_{A B}}{a_{A B}} = 8 \sec$
Magnitude of relative velocity when they are at same position in uAB = $8 \sqrt{2}$ m/s

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