1. ## Kayak paddling across shore

A person is paddling a kayak in a river with a current of 1 ft/s. The kayakers is aimed at the far shore, perpendicular to the current. The kayak's speed in still water would be 4 ft/s. Find the kayak's actual speed and the angle between the kayak's direction and the far shore.

This problem seems easy enough but i feel like i am doing something wrong..

A right triangle would form from (0,0) and (1,4)

So, the hypotenuse would be $\displaystyle \sqrt(17)$

Would this be the speed?

Then the direction would just be

$\displaystyle tan^{-1}(\frac{4}{1})$

so the angle would be 75.96 degrees North of East

Thanks for any input

2. Originally Posted by mybrohshi5
A person is paddling a kayak in a river with a current of 1 ft/s. The kayakers is aimed at the far shore, perpendicular to the current. The kayak's speed in still water would be 4 ft/s. Find the kayak's actual speed and the angle between the kayak's direction and the far shore.

This problem seems easy enough but i feel like i am doing something wrong..

A right triangle would form from (0,0) and (1,4)

So, the hypotenuse would be $\displaystyle \sqrt(17)$

Would this be the speed?
You are probably more comfortable with triangles where the lengths are actual distances and not speeds- so think in terms of one second. If there were no current, in one second, at 4 ft per sec, the kayaker would have gone 4 ft across the river- his new postition would be "(4, 0)". If the kayaker were not paddling but just floating on the current, at 1 ft per sec, the kayaker would have gone 1 foot down river- his new position would be "(0, 1)". Key theoretical point: since velocity is a vector we can just add "components". With both paddling and current, the kayaker goes on the straight line from (0, 0) to (4, 1) which has a length of $\displaystyle \sqrt{17}$ in one second. Yes, his speed would be $\displaystyle \sqrt{17}$ feet per second (You did not include the units- that is an important part of the answer!).

A right triangle with "opposite side" of length 1 and "near side" 4 has angle $\displaystyle tan^{-1}(1/4)= 14$ degrees but that angle is measured to the triangle side across. You correct that the angle the kayak makes with the opposite side is 90- 14= 76 or $\displaystyle tan^{-1}(4/1)= 76$ degrees. But I don't see where you got that "North of East". There is nothing said in the problem, as you posted it, about the direction of the river.

Then the direction would just be

$\displaystyle tan^{-1}(\frac{4}{1})$

so the angle would be 75.96 degrees North of East

Thanks for any input

3. Yeah i always forget to type units when i post here but i always make sure to include them on my paper

and sorry about the North of East. I drew a diagram of the river with a coordinate of North being up and East being to the right but you were right it isnt specified in the question so i will probably just say 76 degrees above the horizontal river bank

Thanks again for clearing this up for me!