A sailboat tacking against the wind moves as follows: 2.6 km at 45° east of north, 4.5 km at 59° west of north, 2 km at 45° east of north. What is the net displacement for the entire motion? Magnitude in km? Direction degrees east of north?
convert each section to its x- and y-components.
Add them up to get the final (x,y) displacement.
Then use the usual Pythagorean formula to get the distance, and
the direction, relative to due East is the angle z such that tan z = x/y
To find the net displacement, we can break down each leg of the sailboat's motion into its horizontal (north-south) and vertical (east-west) components.
First, let's calculate the horizontal and vertical components for each leg:
Leg 1:
- Distance = 2.6 km
- Angle = 45° east of north
Horizontal component = 2.6 km * sin(45°) = 1.84 km east
Vertical component = 2.6 km * cos(45°) = 1.84 km north
Leg 2:
- Distance = 4.5 km
- Angle = 59° west of north (which is the same as 180° - 59° = 121° east of north)
Horizontal component = 4.5 km * sin(121°) = -3.97 km east (negative because it's in the opposite direction)
Vertical component = 4.5 km * cos(121°) = 2.61 km north
Leg 3:
- Distance = 2 km
- Angle = 45° east of north
Horizontal component = 2 km * sin(45°) = 1.41 km east
Vertical component = 2 km * cos(45°) = 1.41 km north
Now, let's add up the horizontal and vertical components separately to find the net horizontal and vertical displacements:
Net horizontal displacement = 1.84 km east - 3.97 km east + 1.41 km east = -0.72 km east
Net vertical displacement = 1.84 km north + 2.61 km north + 1.41 km north = 5.86 km north
The net displacement is the vector sum of the horizontal and vertical displacements:
Net displacement = √(net horizontal displacement^2 + net vertical displacement^2)
Plugging in the values:
Net displacement = √((-0.72 km)^2 + (5.86 km)^2)
Net displacement ≈ √(0.52 km^2 + 34.33 km^2)
Net displacement ≈ √34.85 km^2
Net displacement ≈ 5.91 km
Therefore, the magnitude of the net displacement for the entire motion is approximately 5.91 km.
To find the direction of the net displacement, we can use the inverse tangent function (tan^(-1)):
Direction = tan^(-1)(net horizontal displacement / net vertical displacement)
Plugging in the values:
Direction = tan^(-1)(-0.72 km / 5.86 km)
Direction ≈ -6.7 degrees
Therefore, the direction of the net displacement is approximately 6.7 degrees east of north.