Answer:
Part a)
[tex]v_{bw} = 2.1 mph[/tex]
Part b)
[tex]t = 0.6 h[/tex]
Part c)
[tex]x = 0.4 mile[/tex]
Part d)
[tex]\theta = 46.6 degree[/tex]
Explanation:
Part a)
Velocity of the boat with respect to water stream is given as
[tex]v_{bw} = v_b + v_w[/tex]
[tex]v_{bw} = (1.60 - 2.20sin25) \hat i + 2.20 cos25\hat j[/tex]
so we have
[tex]v_{bw} = 0.67 \hat i + 2 \hat j[/tex]
magnitude of the speed is given as
[tex]v_{bw} = \sqrt{0.67^2 + 2^2}[/tex]
[tex]v_{bw} = 2.1 mph[/tex]
Part b)
Time to cross the river is given as
[tex]t = \frac{y}{v_y}[/tex]
[tex]t = \frac{1.20}{2}[/tex]
[tex]t = 0.6 h[/tex]
Part c)
Distance moved by the boat in downstream is given as
[tex]x = v_x t[/tex]
[tex]x = 0.67 \times 0.6[/tex]
[tex]x = 0.4 mile[/tex]
Part d)
In order to go straight we must net speed along the stream must be zero
so we will have
[tex]vsin\theta = v_w[/tex]
[tex]2.20 sin\theta = 1.60[/tex]
[tex]\theta = 46.6 degree[/tex]