A river of width $200\mathrm{m}$ is flowing from west to east with a speed of $18\mathrm{km/h}$. A boat, moving with speed of $36\mathrm{km/h}$ in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ______ and ______ respectively.

Question

Accepted Answer

C. 1. Convert the given speeds into SI units.

$$
   𝑣_𝑟
   =18	imes\frac{5}{18}

=5\mathrm{m/s},
   $$

where $𝑣_𝑟$ is the speed of the river.

The speed of the boat in still water is

$$
   𝑣_𝑏
   = 36	imes\frac{5}{18}

=10\mathrm{m/s}.
   $$

2. The width of the river is

$$
   𝑤=200,\mathrm{m}.
   $$

The time required to cross the river depends only on the component of the boat's velocity **perpendicular** to the banks.

3. The crossing time is

$$
   𝑡=\frac{𝑤}{𝑣_𝑏\sin	heta},
   $$

where $	heta$ is the angle made by the boat with the bank.

The time is minimum when the boat moves **straight across the river**, i.e.,

$$
   	heta=90^\circ.
   $$

4. Hence, the minimum time for one crossing is

$$
   𝑡_{1ightarrow2}
   = \frac{200}{10}

20\mathrm{s}.
   $$

Similarly, the return trip also takes

$$
   𝑡_{2ightarrow1}
   =

20\mathrm{s}.
   $$

5. Therefore, the total minimum time is

$$
   𝑇
   = 20+20

40\mathrm{s}.
   $$

6. During each crossing, the river carries the boat downstream.

The downstream drift in one crossing is

$$
   	ext{Drift}
   = 𝑣_𝑟	imes 𝑡

= 5	imes20

=100\mathrm{m}.
   $$

7. Since the river flows in the same direction during both crossings, the total displacement along the bank is

$$
   100+100
   =

200\mathrm{m}.
   $$

8. Therefore, the minimum time and displacement along the river bank are

$$
   \boxed{40\mathrm{s}	ext{ and }200\mathrm{m}}
   $$

Hence, the correct option is **C**.