River Boat Vector Physics Problem (Find X and Y Velocity, Angle and Time to Go Straight Across)

You are in a kayak trying to go directly across a river with flow rate of 1 m/s. At what velocity do you travel across the river and at what angle do you paddle at if you are able to kayak at a speed of 2 m/s? How much time will it take to cross the 100 meter wide river?
The first step of this problem is to make the vector of the boat into a right triangle with hypotenuse being the boats velocity. We can then break it up into x and y vectors.
Now if we want to travel at zero velocity up the river. We must have the x vector cancel out the river flow rate. This means that the river flow rate minus the kayak x vector equals zero. Rearanging the formula we get that the river flow rate must equal the kayak x vector.
Now we can solve the right trinangle by using the pythagreon theorem. Which is a squared plus b squared equals c squared.
Rearanging it and pluggin in out numbers we get that the square root of the boats velocity of 2 meters per second squared minus the x velocity of 1 m/s squared equals
1.73 meters per second.
Now we can use soh cah toa to solve for the angle the boat needs to be at to cross the river. We were given the side opposite of the angle and the hypotenuse of the trinangle so lets use the inverse sine of 1 m/s over 2 meters per second. If it is unclear why we are using these values review soh cah toa.
Now we can solve for how long it will take to cross the river. We can take the distance over the velocity perpendicular to the flow in this case y velocity to get the time it takes to cross. Once we plug in 100 meters and 1.73 meters per second we get a time of 57.8 seconds.

Questions?

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