Rain is falling vertically with a speed of $30\; m/s$. $A$ woman rides a bicycle with a speed of $10\; m/s$ in the north to south direction. What is the direction in which she should hold her umbrella?

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

Rain is falling vertically with a speed of $30\,ms^{-1}$. $A$ woman rides a bicycle with a speed of $12\,ms^{-1}$ in the east to west direction. She should hold her umbrella:

An airplane airspeed indicator reads $100 \, m/s$ and its compass shows that it is heading $37^{\circ}$ east of north. The meteorological information provided to the navigator is that the wind velocity is $20 \, m/s$ towards east. The speed of the airplane relative to the ground is closest to ............ $\, m/s$.

$A$ person is swimming with a speed of $10\, m/s$ at an angle of $120^{\circ}$ with the flow of the river and reaches a point directly opposite on the other side. If the speed of the river flow is $x\, m/s$,find the value of $x$ to the nearest integer.

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: