Question

Two particles $A$ and $B$ moving in x-y plane are at origin at $t=0$ second. The initial velocity vectors of $A$ and $B$ are $\overrightarrow{u_A}=8\hat{i}\frac{m}{s}\text{and}\overrightarrow{u_B}=8\hat{j}\frac{m}{s}$. The acceleration of $A$ and $B$ are constant and are $\overrightarrow{a_A}=-2\hat{i}\frac{m}{s^2}\text{and}\overrightarrow{a_B}=-2\hat{j}\frac{m}{s^2}$.
Column I gives certain statements regarding particles $A$ and $B$. Column II gives corresponding results. Match the statements in column I with corresponding results in Column II.
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(A)The time (in seconds) at which velocity}& \left(p\right)16\sqrt{2}\\ \text{of A relative to B is zero}& \\ \text{(B)The distance (in metres) between A and B}& \left(q\right)8\sqrt{2}\\ \text{when their relative velocity is zero}& \\ \text{(C)The time (in seconds) after t = 0 second}& \left(r\right)8\\ \text{at which A and B are at same position}& \\ \text{(D)The magnitude of relative velocity(in}m{s}^{-1}\text{) of A and B}& \left(s\right)4\\ \text{at the instant they are at same position}& \end{array}$

• A. $A-p,B-q,C-p,D-s$
• B. $A-q,B-r,C-s,D-p$
• C. $A-s,B-p,C-r,D-q$
• D. $A-r,B-q,C-p,D-s$

Answer 1

The correct option is C $A-s,B-p,C-r,D-q$

From $\overrightarrow{V}=\overrightarrow{u}+\overrightarrow{a}t$

We have
$\overrightarrow{V_A}=8\hat{i}-2t\hat{i}$
$\overrightarrow{V_B}=8\hat{j}-2t\hat{j}$
Relative velocity of $A$ wrt $B$ is
$\overrightarrow{V_{AB}}=(8-2t)\hat{i}-(8-2t)\hat{j}$
$\overrightarrow{V_{AB}}=0\Rightarrow 8-2t=0$
$\Rightarrow t=4s$

Position of $A$ wrt $B$:
$\overrightarrow{r_{AB}}=\overrightarrow{U_{AB}}t+\frac{1}{2}\overrightarrow{a_{AB}}{t}^2$
$\overrightarrow{r_{AB}}=(8\hat{i}-8\hat{j})t+\frac{1}{2}(-2\hat{i}+2\hat{j})t^2\overrightarrow{r_{AB}}=(8t-t^2)\hat{i}-(8t-t^2)\hat{j}$

When $\overrightarrow{V_{AB}}=0,t=4$. So,
$\overrightarrow{r_{AB}}$${|}_{t=4}$ $(32-16)\hat{i}-(32-16)\hat{j}$
$\overrightarrow{r_{AB}}=16\hat{i}-16\hat{j}$
$|\overrightarrow{r_{AB}}|=16\sqrt{2}m$

When particles are at same position
$\overrightarrow{r_{AB}}=0\Rightarrow 8t-t^2=0$
$\Rightarrow t=8s$
$\overrightarrow{V_{AB}}$|${}_{t=8}$ = $-8\hat{i}-8\hat{j}$
$|\overrightarrow{V_{AB}}|=8\sqrt{2}m{s}^{-1}$