1. The problem statement, all variables and given/known data A kayaker needs to paddle north across a 105 m wide harbor. The tide is going out, creating a tidal current that flows to the east at 1.5 m/s. The kayaker can paddle with a speed of 3.4 m/s. (a) In which direction should he paddle in order to travel straight across the harbor? (degrees west of north) (b) How long will it take him to cross? (seconds) 2. Relevant equations c^2= a^2 + b^2 3. The attempt at a solution I divided 105m by 3.4 m/s to determine how long it would take to go straight across the harbor. Then I multiplied that number (30.88) by 1.5 m/s to get the distance he would have ended up downstream. 46.32m. Then, I attempted to set up a right triangle and solve for the angle but that's where I got confused.
The kayaker will take longer than [30.88] that to cross the harbour, since he/she is not paddling directly towards the opposite side.
1. The problem statement, all variables and given/known data A kayaker needs to paddle north across a 105 m wide harbor. The tide is going out, creating a tidal current that flows to the east at 1.5 m/s. The kayaker can paddle with a speed of 3.4 m/s. (a) In which direction should he paddle in order to travel straight across the harbor? (degrees west of north) (b) How long will it take him to cross? (seconds) Then, I attempted to set up a right triangle and solve for the angle but that's where I got confused. ----------------------------------------------------- Yes you can use a right triangle to solve the problem. This is a vector problem and using a right triangle geometry is one of the methods. Kayaker speed with direction is one vector. The tide speed and direction is another vector. You can do operations on this 2 vectors and in this example adding the two vectors. The sum of the two vectors will result in the kayaker paddling straight across the habour.