Question

A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 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 the opposite bank?

Answer 1

Given:
Distance between the opposite shore of the river or width of the river = 400 m
Rate of flow of the river = 2.0 m/s
Boat is sailing at the rate of 10 m/s.
The vertical component of velocity 10 m/s takes the boat to the opposite shore. The boat sails at the resultant velocity v_r.
Time taken by the boat to reach the opposite shore:

$$\text{ Time }=\frac{\text{ Distance }}{\text{ Time }}=\frac{400}{10}=40s$$

From the figure, we have:

$$\tan \theta =\frac{2}{10}=\frac{1}{5}$$

The boat will reach point C.

$$\text{ In }∆ABC,$$

$$\tan \theta =\frac{BC}{AB}=\frac{BC}{400}=\frac{1}{5}$$

$$\Rightarrow BC=\frac{400}{5}=80\text{ m }$$

Magnitude of velocity

$$\left|v_r\right|=\sqrt{{10}^2+2^2}=10.2\text{ m/s }$$

Let α be the angle made by the boat sailing with respect to the direction of flow.

$$\tan \left(\alpha \right)=\frac{10}{2}$$

$$\Rightarrow \alpha =78.7^{\circ }$$

Distance the boat need to travel to reach the opposite shore = $$\frac{400}{\sin \left(\alpha \right)}=407.9\text{ m }$$ 

 Using Pythagoras' theorem, we get:
Distance = $$\sqrt{407.9^2-{400}^2}=79.9\text{ m }\approx 80\text{ m}$$