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):
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):
where θ is the angle of the boat with respect to the shoreline (upstream). Therefore,
Her actual speed across the river is her boat's other velocity component, which follows from the definition of sine:
vacross=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!