A boat crosses a river, 200 m wide, in minimum possible time. If velocity of river is 5 m/s and velocity of boat is still water is 10 m/s. Then, find time taken to cross the river and displacement of the boat.

The correct option is 20 sec. and 100√5m.

Step 1: Understand the condition for minimum time
To cross a river of width $d$ in the minimum possible time, the boat must be steered directly perpendicular to the river flow (i.e., directed straight across the river bank). In this orientation, the velocity component of the boat perpendicular to the river bank is equal to the full speed of the boat in still water, $v_b$.

Given parameters:
Width of the river, $d=200\text{ m}$
Velocity of the boat in still water, $v_b=10\text{ m/s}$
Velocity of the river current, $v_r=5\text{ m/s}$

Step 2: Calculate the time taken to cross the river
The time taken ($t$) to cross the river is given by the ratio of the river width to the velocity of the boat perpendicular to the banks:

$t=\frac{d}{v_b}$

Substitute the given values:

$t=\frac{200}{10}=20\text{ sec}$

Step 3: Calculate the drift of the boat
While crossing, the river flow carries the boat downstream along the direction of flow. This downstream distance (drift, $x$) covered in time $t$ is calculated as:

$x=v_r\times t$

Substitute the values:

$x=5\times 20=100\text{ m}$

Step 4: Calculate the total net displacement of the boat
The total displacement ($S$) is the vector resultant of the perpendicular motion across the river ($d$) and the downstream drift ($x$). Using the Pythagorean theorem:

$S=\sqrt{d^2+x^2}$

Substitute $d=200$ and $x=100$:

$S=\sqrt{{200}^2+{100}^2}=\sqrt{40000+10000}=\sqrt{50000}$

Simplifying the square root:

$S=100\sqrt{5}\text{ m}$

Thus, the time taken to cross the river is 20 sec. and the net displacement of the boat is 100√5m.