Suppose a ship heads directly north away from the docks in a harbor at a speed of 3.80 km/h. after traveling 2.50 km, the ship leaves the harbor encountering an ocean current moving due west at 3.00 km/h. find the location of the ship (distance from the dock in km and direction in degrees counterclockwise from the east) 1.00 h after it leaves the dock if it does not change its direction relative to the ocean.

   R = 3,936 km      θ=105º

Explanation:

This is a problem of vector addition, let's find it takes time for the ship to reach the ocean

      v1 = x1 / t1

      t1 = 2.50 / 3.80

      t1 = 0.658 h

Let's analyze how much travel time is left

       t2 = 1 h - t1

       t2 = 1 - 0.658

       t2 = 0.342 h

This is the time he walked by the ocean, let's calculate every distance he traveled

X axis

          v2 = -3.00 km / h

          x = v t

          x2 = -3.00 * 0.342

          x2 = - 1,026 km

Axis y

         v1 = 3.80 km / h

         y2 = v1 t2

         y2 = 3.80 0.342

         y2 = 1.30 km

The total distance on each axis is

         xall = x2

         xall = -1,026 km

         y all = 2.5 + y2

        y  all = 2.5 + 1.3

        y all = 3.8 km

Let's calculate the distance and angle from the pier with the Pythagorean theorem and trigonometry

       R² = xall² + yall²

       R = √(1,026² + 3.8²)

       R = 3,936 km

       tan θ = yall / xall

       tan θ = 3.8 / 1.026

        θ = tan -1 (3.70)

      θ = 75º

The angle measured from the x axis (East) is 180 - 74 = 105º

it has to be b, about 97 percent.