The speed of a swimmer is $4\,km\,h^{-1}$ in still water. If the swimmer makes his strokes normal to the flow of a river of width $1\,km$,he reaches a point $750\,m$ downstream on the opposite bank. The speed of the river water is $.........\,km\,h^{-1}$.

$A$ swimmer wants to cross a river from point $A$ to point $B$. Line $AB$ makes an angle of $30^{\circ}$ with the flow of the river. The magnitude of the velocity of the swimmer is the same as that of the river. The angle $\theta$ with the line $AB$ should be in degrees,so that the swimmer reaches point $B$.

$A$ particle is moving along the $x$-axis with its coordinate with time '$t$' given by $x(t) = 10 + 8t - 3t^2$. Another particle is moving along the $y$-axis with its coordinate as a function of time given by $y(t) = 5 - 8t^3$. At $t = 1 \; s$,the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$. Then $v$ (in $m^2/s^2$) is:

$A$ river is flowing from $W$ to $E$ with a speed of $5 \, m/min$. $A$ man can swim in still water with a velocity of $10 \, m/min$. In which direction should the man swim to take the shortest possible path to reach the opposite bank (South)?

$A$ boat is moving with a velocity of $3\hat i + 4\hat j$ in a river,and the water is moving with a velocity of $-3\hat i - 4\hat j$ with respect to the ground. The relative velocity of the boat with respect to the water is:

$A$ stationary man observes that the rain is falling vertically downward. When he starts running with a velocity of $12 \ km/h$,he observes that the rain is falling at an angle of $60^{\circ}$ with the vertical. The actual velocity of the rain is: