Two trains, each $30 \,m$ long, are travelling in opposite directions with velocities $5 \,m/s$ and $10 \,m/s$. They will cross each other after: (in $\,s$)

Two girls are standing at the ends $A$ and $B$ of a ground where $AB = b$. The girl at $B$ starts running in a direction perpendicular to $AB$ with velocity $V_1$. The girl at $A$ starts running simultaneously with velocity $V_2$ and meets the other girl in the shortest possible time $t$. The value of $t$ is:

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:

$A$ river $200 \,m$ wide is flowing at a rate of $3.0 \,m/s$. $A$ boat is sailing at a velocity of $15 \,m/s$ with respect to the water in a direction perpendicular to the river. How far from the point directly opposite to the starting point does the boat reach on the opposite bank (in $\,m$)?

If a $100 \,m$ long train needs $7.2 \,s$ to cross an object moving in a direction opposite to the train's direction with a speed of $5 \,km/h$, then find the velocity of the train. (in $\,km/h$)

$A$ slider block $A$ moves downward at a speed of $v_A = 2 \ m/s$,at an angle of $75^\circ$ with the horizontal as shown in the figure. The velocity of the portion of belt $B$ between ideal pulleys $C$ and $D$ with respect to $A$ is $v_{CD/A} = 2 \ m/s$ at an angle $\theta$ with the horizontal. The magnitude of the velocity of portion $CD$ of the belt when $\theta = 15^\circ$ is .......... $m/s$.