A ship leaves Team harbor at 10:00am one morning. The captain steers a course bearing 136 degrees at a speed of 20km/h through the water. There is a current of 5km/h in a direction of 046 degrees. Find
1. The direction in which the boat travels.
2. The distance from Team at 10:00am
If you check this out, you will see that the velocity vectors form a right triangle, making things pretty easy.
The final speed is just the hypotenuse of the triangle: √(20^2+5^2) = √425 ≈ 20.6 km/hr
The direction is just (136-θ)° where tanθ = 5/20
the distance is of course 0 km, since no time has passed. I feel a typo coming on ...