Answer:
a. 25.98i - 15j mi/h
b. 1.71i + 4.7j mi/h
c. 27.69i -10.3j mi/h
Step-by-step explanation:
a. Identify the ship's vector
Since the ship moves through water at 30 miles per hour at an angle of 30° south of east, which is in the fourth quadrant. So, the x-component of the ship's velocity is v₁ = 30cos30° = 25.98 mi/h and the y-component of the ship's velocity is v₂ = -30sin30° = -15 mi/h
Thus the ship's velocity vector as ship moves through the water v = v₁i + v₂j = 25.98i + (-15)j = 25.98i - 15j mi/h
b. Identify the water current's vector
Also, since the water is moving at 5 miles per hour at an angle of 20° south of east, this implies that it is moving at an angle 90° - 20° = 70° east of north, which is in the first quadrant. So, the x-component of the water's velocity is v₃ = 5cos70° = 1.71 mi/h and the y-component of the water's velocity is v₄ = 5sin70° = 4.7 mi/h
Thus the ship's velocity vector v' = v₃i + v₄j = 1.71i + 4.7j mi/h
c. Identify the vector representing the ship's actual motion.
The velocity vector representing the ship's actual motion is V = velocity vector of ship as ship moves through water + velocity vector of water current.
V = v + v'
= 25.98i - 15j mi/h + 1.71i + 4.7j mi/h
= (25.98i + 1.71i + 4.7j - 15j )mi/h
= 27.68i -10.3j mi/h
