Question

HELPP PPLEASEEE!!!

A ship moves through the water at 30 miles/hour at an angle of 30° south of east. The water is moving 5 miles/hour at an angle of 20° east of north. Identify the ship's vector, the water current's vector, and the vector representing the ship's actual motion.

Answer 1

The ship's vector representing its actual motion is 30.73 mph east of north.

Step-by-step explanation:

To solve this problem, we can break down the velocities of the ship and the water current into their horizontal and vertical components. The ship's vector can be represented as:

Ship's Vector: 30 mph at an angle of 30° south of east

Breaking this down into horizontal and vertical components:

Horizontal Component = 30 mph * cos(30°) = 25.98 mph east

Vertical Component = 30 mph * sin(30°) = 15 mph south

The water current's vector can be represented as:

Water Current's Vector: 5 mph at an angle of 20° east of north

Breaking this down into horizontal and vertical components:

Horizontal Component = 5 mph * cos(20°) = 4.75 mph north

Vertical Component = 5 mph * sin(20°) = 1.71 mph east

To find the ship's actual motion, we can add the horizontal and vertical components together:

Horizontal Component = 25.98 mph east + 4.75 mph north = 30.73 mph east of north

Vertical Component = 15 mph south + 1.71 mph east = 16.71 mph south of east

Therefore, the ship's vector representing its actual motion is 30.73 mph east of north.

Answer 2

See below in bold.

Step-by-step explanation:

Ship's vector:

Horizontal component = 30 cos 30 = 25.98.

Vertical component = 30 sin(-30) = -15.

So it is <25.98, -15).

The current's vector:

Horizontal component = 5 sin 20 = 1.71.

Vertical component = 5 cos 20 = 4.7.

So it is <1.71, 4.7>.