A 200-m-wide river has a uniform flow speed of 0.99 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.4 m/s with respect to the water. There is a clearing on the north bank 35 m upstream from a point directly opposite the clearing on the south bank.

a. At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank?
b. How long will the boat take to cross the river and land in the clearing?

Let [tex]v_v[/tex] be the vertical component of the boat velocity with respect the the river, pointing North. Let [tex]v_h[/tex] be the horizontal component of the boat velocity with respect to the river, pointing West, aka upstream. Since the total velocity of the boat is 4.4m/s

[tex]v_v^2 + v_h^2 = 4.4^2 = 19.36[/tex]

The time it takes for the boat to cross 200m-wide river at [tex]v_v[/tex] rate is

[tex]t = 200 / v_v[/tex] or [tex]v_v = 200 / t[/tex]

This is also the time it takes for the boat to travel 35m upstream, horizontally, at the rate of [tex] v_h - 0.99[/tex] m/s

[tex]t = \frac{35}{v_h - 0.99}[/tex]

[tex]v_h - 0.99 = 35/t[/tex]

[tex]v_h = 35/t + 0.99[/tex]

We can substitute [tex]v_v,v_h[/tex] into the total velocity equation to solve for t

[tex]\frac{200^2}{t^2} + (\frac{35}{t} + 0.99)^2 = 19.36[/tex]

[tex]\frac{40000}{t^2} + \frac{35^2}{t^2} + 2*0.99*\frac{35}{t} + 0.99^2 = 19.36[/tex]

From here we can multiply both sides by [tex]t^2[/tex]

[tex]40000 + 1225 + 69.3t + 0.9801t^2 = 19.36t^2[/tex]

[tex]18.38 t^2 - 69.3t - 41225 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{69.3\pm \sqrt{(-69.3)^2 - 4*(18.3799)*(-41225)}}{2*(18.38)}[/tex]

[tex]t= \frac{69.3\pm1742.31}{36.7598}[/tex]

t = 49.28 or t = -45.51

Since t can only be positive we will pick t = 49.28

[tex]v_h = 35 / t + 0.99 = 35 / 49.38 + 0.99 = 1.7 m/s[/tex]

The angle, relative to the flow of river direction is

[tex]cos(\alpha) = \frac{v_h}{v} = \frac{1.7}{4.4} = 0.3864[/tex]

[tex]\alpha = cos^{-1}(0.3864) = 1.174 rad[/tex] or 67.3 degree