Answer:
a) 10.29° upstream
b) t=338.7s
Explanation:
If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:
[tex]\alpha =atan(\frac{0.5}{1})=26.56°[/tex]
[tex]D=\sqrt{1^2+0.5^2}=1.118km[/tex]
The realitve velocity of the boat respect to the water is:
[tex]V_{B/W}=[3*cos\beta ,3*sin\beta ][/tex] where β is the angle it has to be pointed at.
From the relative mvement equations:
[tex]V_{B/W}=V_B-V_W[/tex] where [tex]V_B=[V*cos\alpha ,-V*sin\alpha ][/tex]
From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:
[tex](3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2[/tex]
[tex]V^2-4*V*sin\alpha -5=0[/tex] Solving for V:
V = 3.3m/s and V=-1.514m/s Replacing this value into one of our previous x or y-axis equations:
[tex]\beta =acos(\frac{V*cos\alpha }{3} ) = 10.29°[/tex]
The amount of time:
[tex]t = D/V = \frac{1118m}{3.3m/s} =338.7s[/tex]
