In part 1, we have two perpendicular velocities due to the kayaker and the river: 2.5 m/s across the river and 1.5 m/s downstream. The magnitude of the resultant velocity follows from the Pythagorean theorem:

v= √(1.5²+2.5²) = 2.9 m/s.

The angle of the resultant velocity (relative to the shoreline) follows from the definition of the tangent function (opposite/adjacent):

θ= tan^{-1}(2.5/1.5) = 59°

To find the time it takes her to cross the river at this angle, we first need the distance d she has to cover, which follows from the definition of the sine function (opposite/hypotenuse):

sinθ=2.0 km/d,

d=2.0km/sin(59°) = 2.33 km =2330 m

Then the time follows from the definition of velocity,

v=d/t, t=d/v = 2330m/2.92m/s = 800 s.

(Notice that we keep 3 significant digits in the calculations, but round the final answer to 2 significant digits.)

In Part II, we want the resultant velocity to be across the river. This means the river's 1.5 m/s downstream velocity must be canceled by the boat's upstream velocity component. This upstream component follows from the definition of cosine (adjacent/hypotenuse):

1.5=2.5cosθ

where θ is the angle of the boat with respect to the shoreline (upstream). Therefore,

θ=cos^{-1}(1.5/2.5)=53.1°

Her actual speed across the river is her boat's other velocity component, which follows from the definition of sine:

v_{across}=2.5m/s sin(53.1) =2.0 m/s.

At this speed, it will take her

t=2000 m/2.0 m/s =1000 s

to cross the river.

Note: the shortest path is not always the fastest!

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