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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 ${\vec{a}}_{B} = - 2 \hat{j} m / s^{2} .$ List I gives certain statements regarding particle A and B. List II gives corresponding results.
List- I List – 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 (in m/s) of
A and B at the instant they are at same position. S) 2
T) 4
Then the correct match of the lists is

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A – S, B – P, C – R, D – Q

A – R, B – P, C – S, D – T

A – T, B – P, C – S, D – R

A – T, B – P, C – R, D – Q

answer is D.

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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 –
${\vec{a}}_{A B} = {\vec{a}}_{A} - {\vec{a}}_{B} = (- 2 \hat{i} + 2 \hat{j}) m / s^{2}$
$∴ a_{A B} = 2 \sqrt{2} m / s^{2}$
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 $V_{A B} = 0$ is $t = \frac{u_{A B}}{a_{A B}} = 4 \sec .$
The distance between A & B when $V_{A B} = 0$ is $S = \frac{{u^{2}}_{A B}}{2 a_{A B}} = 16 \sqrt{2} m$
The time when both are at same position is -
$T = \frac{2 u_{A B}}{a_{A B}} = 8 \sec .$
Magnitude of relative velocity when they are at same position in $u_{A B} = 8 \sqrt{2} m / s .$

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