(A) Let at time $t$, the position of $A$ be $\overrightarrow{r}_A = (0 + 0t)\hat{i} + (0 + 2t)\hat{j} = 2t\hat{j}$.Let at time $t$, the position of $B

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(A) Let at time $t$, the position of $A$ be $\overrightarrow{r}_A = (0 + 0t)\hat{i} + (0 + 2t)\hat{j} = 2t\hat{j}$.Let at time $t$, the position of $B$ be $\overrightarrow{r}_B = (0 + 2t)\hat{i} + (10 + 0t)\hat{j} = 2t\hat{i} + 10\hat{j}$.The relative position vector $\overrightarrow{r}_{AB} = \overrightarrow{r}_B - \overrightarrow{r}_A = 2t\hat{i} + (10 - 2t)\hat{j}$.The square of the distance $D^2 = |\overrightarrow{r}_{AB}|^2 = (2t)^2 + (10 - 2t)^2$.$D^2 = 4t^2 + 100 + 4t^2 - 40t = 8t^2 - 40t + 100$.For minimum distance, $\frac{d(D^2)}{dt} = 0$.$16t - 40 = 0$.$t = \frac{40}{16} = 2.5 \text{ s}$.Since $\frac{d^2(D^2)}{dt^2} = 16 > 0$, the distance is minimum at $t = 2.5 \text{ s}$.

Class 11 Physics 3-2.Motion in Plane Relative Velocity (river boat, rain, wind)

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Two persons $A$ and $B$ are located in the $X-Y$ plane at the points $(0,0)$ and $(0,10)$ respectively. (The distances are measured in $MKS$ units). At time $t=0$, they start moving simultaneously with velocities $\overrightarrow{v}_A = 2\hat{j} \text{ m/s}$ and $\overrightarrow{v}_B = 2\hat{i} \text{ m/s}$ respectively. The time after which $A$ and $B$ are at their closest distance is:

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A
$2.5$ s
B
$4$ s
C
$1$ s
D
$\frac{10}{\sqrt{2}} \text{ s}$

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3-2.Motion in Plane

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Relative Velocity (river boat, rain, wind)

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Two particles having position vectors $\overrightarrow{r_1} = (3\hat{i} + 5\hat{j}) \text{ m}$ and $\overrightarrow{r_2} = (-5\hat{i} - 3\hat{j}) \text{ m}$ are moving with velocities $\overrightarrow{v_1} = (4\hat{i} + 3\hat{j}) \text{ m/s}$ and $\overrightarrow{v_2} = (\alpha\hat{i} + 7\hat{j}) \text{ m/s}$. If they collide after $2 \text{ s}$,the value of $\alpha$ is:

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$A$ swimmer can swim with speed '$v$' with respect to still water in a river which is flowing with speed '$u$'. There is a float moving with the river. Now the swimmer overtakes the float,gets a lead of '$l$',and returns back to the float. The time taken by the swimmer in this process will be:

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$A$ man of mass $60 \,kg$ is standing on a boat of mass $140 \,kg$,which is at rest in still water. The man is initially at $20 \,m$ from the shore. He starts walking on the boat for $4 \,s$ with a constant speed of $1.5 \,m/s$ towards the shore. The final distance of the man from the shore is .......... $m$.

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Match Column-$I$ with Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Velocity of $A$ relative to $B$ if $A$ and $B$ are moving perpendicular to each other $(a)$ $\vec{v}_{rm} = \vec{v}_r + \vec{v}_m$

$(2)$ Velocity of rain drops relative to a man $(b)$ $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$

$(c)$ $v_{AB} = \sqrt{v_A^2 + v_B^2}$

Column-$I$	Column-$II$
$(1)$ Velocity of $A$ relative to $B$ if $A$ and $B$ are moving perpendicular to each other	$(a)$ $\vec{v}_{rm} = \vec{v}_r + \vec{v}_m$
$(2)$ Velocity of rain drops relative to a man	$(b)$ $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$
	$(c)$ $v_{AB} = \sqrt{v_A^2 + v_B^2}$

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$A$ $120 \, m$ long train is moving towards west with a speed of $10 \, m/s$. $A$ bird flying towards east with a speed of $5 \, m/s$ crosses the train. The time taken by the bird to cross the train will be ........ $sec$.

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$A$ swimmer can swim with speed '$v$' with respect to still water in a river which is flowing with speed '$u$'. There is a float moving with the river. Now the swimmer overtakes the float,gets a lead of '$l$',and returns back to the float. The time taken by the swimmer in this process will be:

$A$ $120 \, m$ long train is moving towards west with a speed of $10 \, m/s$. $A$ bird flying towards east with a speed of $5 \, m/s$ crosses the train. The time taken by the bird to cross the train will be ........ $sec$.

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