Question

A person in a canoe can paddle his canoe at a steady 3.0 m/s in still water. He wishes to cross a 2.4 km wide river that has a current of 1.8 m/s traveling in the northward direction. Give step by step details in your solutions and solve graphically and analytically.

A) If this person first aims his canoe straight across the river in the eastern direction, the current will carry him downstream as he paddles across. What will be his actual velocity (magnitude and angle with respect to the eastward diredction) as he crosses? How long will it take him to cross the river?

B) If he aims the canoe somewhat upstream, he can actually travel straight across the river. In what direction must he aim? What is his actual speed across the river for this situation, and how long will it take him to cross? (Hint: The downstream direction found in part (a) will not be the same as the upstream angle required in part b!) (Hint: You will need a compass for the graphical method)

Answer 1

In part 1, we have two perpendicular velocities due to the kayaker and the river: 3m/s across the river and 1.8 m/s downstream. The magnitude of the resultant velocity follows from the Pythagorean theorem:

v= √(1.8²+3²) = 3.498 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(3/1.8) = 59.03°

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 = 2.4 km/d,

d=2.4km/sin(59.03°) = 2.799km =2799 m

Then the time follows from the definition of velocity,

v=d/t, t=d/v = 2799m/3.498m/s = 800.171 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 1.8 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):

1.8=3cosθ

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

θ=cos^-1(1.8/3)=53.13°

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

v_across=3m/s sin(53.13) =2.399 m/s.

At this speed, it will take her

t=2400m/2.399 m/s =1000 s to cross the river.

Note: the shortest path is not always the fastest!

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