$A$ man goes $10\,km$ downstream in $2\,hrs$ and $30\,km$ upstream in $10\,hrs$. How much time will he take to swim $40\,km$ in still water?

Two parallel rail tracks run north-south. Train $A$ moves north with a speed of $54\; km\; h^{-1},$ and train $B$ moves south with a speed of $90\; km\; h^{-1}.$ What is the velocity (in $m\; s^{-1}$) of a monkey running on the roof of the train $A$ against its motion (with a velocity of $18\; km\; h^{-1}$ with respect to the train $A$) as observed by a man standing on the ground?

$A$ river has a steady speed of $v$. $A$ man swims upstream for a distance of $d$ and swims back to the starting point in a total time $t$. The man can swim at a speed of $2v$ in still water. If the time taken by the man in still water is $t_0$ to complete the same total distance of $2d$, then $\frac{t}{t_0}$ is

$A$ man walking briskly in rain with speed $v$ must slant his umbrella forward making an angle $\theta$ with the vertical. $A$ student derives the following relation between $\theta$ and $v: \tan \theta = v$ and checks that the relation has a correct limit: as $v \rightarrow 0, \theta \rightarrow 0,$ as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not,guess the correct relation.

$A$ bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of $54 \text{ kmh}^{-1}$ while the other has a speed of $36 \text{ kmh}^{-1}$. The bird starts moving from the first car towards the other and is moving with a speed of $36 \text{ kmh}^{-1}$ when the two cars are separated by $36 \text{ km}$. The total distance covered by the bird before the cars meet each other is: (in $\text{ m}$)

$A$ man sitting in a bus travelling from west to east with a speed of $40 \, km/h$ observes that the rain-drops are falling vertically down. To another man standing on the ground,the rain will appear: