An observer moves with a constant speed along the line joining two stationary objects. He will observe that the two objects

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:

An aeroplane is to go along a straight line from $A$ to $B$,and back again. The relative speed with respect to wind is $V$. The wind blows perpendicular to the line $AB$ with speed $v$. The distance between $A$ and $B$ is $\ell$. The total time for the round trip is:

Rain is falling vertically with a speed of $35 \; m s^{-1}$. $A$ woman rides a bicycle with a speed of $12 \; m s^{-1}$ in the east to west direction. What is the direction in which she should hold her umbrella?

$A$ river is flowing due east with a speed $3\, ms^{-1}$. $A$ swimmer can swim in still water at a speed of $4\, ms^{-1}$ (figure).
$(a)$ If the swimmer starts swimming due north,what will be his resultant velocity (magnitude and direction)?
$(b)$ If he wants to start from point $A$ on the south bank and reach the opposite point $B$ on the north bank,
$(i)$ In which direction should he swim?
$(ii)$ What will be his resultant speed?
$(c)$ From the two different cases as mentioned in $(a)$ and $(b)$ above,in which case will he reach the opposite bank in a shorter time?

$A$ boy runs on a horizontal road with a speed of $4 \ m \ s^{-1}$ while it is raining. He sees that the rain is making an angle $\theta$ with the vertical while running from west to east. However, when he runs from east to west, the angle is $\alpha$. The rain is pouring down at an angle $45^{\circ}$ with the vertical and at a speed of $8 \ m \ s^{-1}$ as shown in the figure. The ratio $\frac{\tan \theta}{\tan \alpha}$ is,