a. 4.4 m/s southeast
The velocity of the boat has two components:
- A component towards east direction, of magnitude [tex]v_x = 3.8 m/s[/tex]
- A component towards the south direction, of magnitude [tex]v_y = 2.2 m/s[/tex]
Since the two components are perpendicular to each other, the magnitude of the resultant can be found by using Pythagorean theorem:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(3.8 m/s)^2+(2.2 m/s)^2}=4.4 m/s[/tex]
and the direction is between the directions of the two components, so southeast.
b. 10.8 s
In this part of the problem, we can just use the velocity of the boat along the direction perpendicular to the river, which is
[tex]v_x = 3.8 m/s[/tex]
the distance between the two sides of the river is
d = 41 m
Therefore, the time taken to cross the river is
[tex]t=\frac{d}{v_x}=\frac{41 m}{3.8 m/s}=10.8 s[/tex]
c. 23.76 m
In order to find how far the boat will land downstream, we have to consider its motion along the direction of the stream, which occurs with a velocity of
[tex]v_y = 2.2 m/s[/tex]
We have calculated previously that the time taken to cross the river is
[tex]t=10.8 s[/tex]
So, the boat will land downstream by a distance of
[tex]d=v_y t = (2.2 m/s)(10.8 s)=23.76 m[/tex]
