Question

Two particles $A$ and $B$ moving in $x-y$ plane are at origin at $t=0\text{sec}$.
The initial velocity vectors of $A$ and $B$ are ${\overrightarrow{u}}_A=8\hat{i}\text{m/s}$ and ${\overrightarrow{u}}_B=8\hat{j}\text{m/s}$.
The acceleration of $A$ and $B$ are constant and are ${\overrightarrow{a}}_A=-2\hat{i}{\text{m/s}}^2$ and ${\overrightarrow{a}}_B=-2\hat{j}{\text{m/s}}^2$.
List-I gives certain statements regarding particle $A$ and $B$.
List - II gives corresponding results in SI units.
Match the statements in List-I with corresponding results in List-II.
$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \text{(I)}& \text{The time (in seconds) at which velocity of A relative}& \text{(P)}& 16\sqrt{2}\\ & \text{to B is zero}& & \\ \text{(II)}& \text{The distance (in metres) between A and B when their}& \text{(Q)}& 8\sqrt{2}\\ & \text{relative velocity is zero}& & \\ \text{(III)}& \text{The magnitude of relative velocity of A and B at the}& \text{(R)}& 8\\ & \text{instant they are at same position}& & \\ & & \text{(S)}& 4\end{array}$

• A. $I\to S;II\to P;III\to Q$
• B. $I\to P;II\to S;III\to Q$
• C. $I\to S;II\to Q;III\to S$
• D. $I\to Q;II\to P;III\to R$

Answer 1

The correct option is A $I\to S;II\to P;III\to Q$

(I)
Velocity of A is ${\overrightarrow{V}}_A=(8\hat{i}-2\hat{i}t)$ Velocity of B is ${\overrightarrow{V}}_B=(8\hat{j}-2\hat{j}t)$
Velocity of A wrt B
${\overrightarrow{V}}_A-{\overrightarrow{V}}_B=0\Rightarrow t=4\text{sec}$

(II)
$S_A=(8t-\frac{1}{2}2t^2)\hat{i}$
At $t=4,S_A=8\times 4\hat{i}-16\hat{i}=+16\hat{i}$
$S_B=(8t-\frac{1}{2}2t^2)\hat{j}$
At $t=4,S_B=8\times 4\hat{i}-16\hat{j}=+16\hat{j}$
$\therefore$ Magnitude of distance between them $=16\sqrt{2}$

(III) $S_A-S_B=0\Rightarrow 8t-\frac{1}{2}2t^2=0\Rightarrow t=8\text{sec}$
$\overrightarrow{V_A}=(8-2t)\hat{i}=-8\hat{i}\text{m/s}$ $\overrightarrow{V_B}=-8\hat{j}\text{m/s}$
$\therefore$ Magnitude of relative velocity at this instant is $8\sqrt{2}\text{m/s}$