1. the pilot of a private plane flies 20km in a direction 60 degrees north of east,then 30km straight east,then 10km straight north.how far and in what direction is the plane from the starting point.
2. two forces acts on a point object as follows 100N at 170 degrees and 100N at 50 degrees.find their resultant.
3. A car is traveling eastwards at a speed of 40m/s.But a 3m/s wind is blowing southward.What are the direction and speed of the car relative to its original direction.
Review how to add vectors.
Please do not dump your homework here. You need to make an effort besides cutting and pasting questions.
1. To find the distance and direction of the plane from the starting point, we can break down the planes' movements into components.
First, let's calculate the eastward (x) and northward (y) components of each leg of the journey:
Leg 1: 20 km at 60 degrees north of east
x1 = 20 km * cos(60 degrees) ≈ 10 km
y1 = 20 km * sin(60 degrees) ≈ 17.32 km
Leg 2: 30 km straight east
x2 = 30 km
y2 = 0 km
Leg 3: 10 km straight north
x3 = 0 km
y3 = 10 km
Now, let's sum up the x and y components to find the total displacement from the starting point:
Total x displacement = x1 + x2 + x3 ≈ 10 km + 30 km + 0 km = 40 km
Total y displacement = y1 + y2 + y3 ≈ 17.32 km + 0 km + 10 km ≈ 27.32 km
To find the distance, we use the Pythagorean theorem:
Distance = sqrt((Total x displacement)^2 + (Total y displacement)^2)
Distance = sqrt((40 km)^2 + (27.32 km)^2)
Distance ≈ sqrt(1600 km^2 + 746.74 km^2)
Distance ≈ sqrt(2346.74 km^2)
Distance ≈ 48.44 km (rounded to two decimal places)
To find the direction, we use trigonometry and inverse tangent (arctan):
Direction = arctan(Total y displacement / Total x displacement)
Direction = arctan(27.32 km / 40 km)
Direction ≈ 35.1 degrees (rounded to one decimal place)
Therefore, the plane is approximately 48.44 km away from the starting point in a direction of 35.1 degrees north of east.
2. To find the resultant of two forces, we can add their x and y components separately.
Force 1: 100 N at 170 degrees
x1 = 100 N * cos(170 degrees) ≈ -97.38 N (negative because it acts in the opposite direction)
y1 = 100 N * sin(170 degrees) ≈ -28.64 N
Force 2: 100 N at 50 degrees
x2 = 100 N * cos(50 degrees) ≈ 64.28 N
y2 = 100 N * sin(50 degrees) ≈ 76.6 N
Now, let's sum up the x and y components of the forces:
Resultant x component = x1 + x2 ≈ -97.38 N + 64.28 N ≈ -33.1 N
Resultant y component = y1 + y2 ≈ -28.64 N + 76.6 N ≈ 47.96 N
To find the magnitude of the resultant force, we use the Pythagorean theorem:
Resultant magnitude = sqrt((Resultant x component)^2 + (Resultant y component)^2)
Resultant magnitude = sqrt((-33.1 N)^2 + (47.96 N)^2)
Resultant magnitude ≈ sqrt(1094.41 N^2 + 2305.44 N^2)
Resultant magnitude ≈ sqrt(3399.85 N^2)
Resultant magnitude ≈ 58.34 N (rounded to two decimal places)
To find the direction, we use trigonometry and inverse tangent (arctan):
Resultant direction = arctan(Resultant y component / Resultant x component)
Resultant direction = arctan(47.96 N / -33.1 N)
Resultant direction ≈ -52.9 degrees (rounded to one decimal place)
Therefore, the resultant of the two forces is approximately 58.34 N at an angle of -52.9 degrees.
3. To find the direction and speed of the car relative to its original direction, we can use vector addition.
Let's call the car's velocity as Vcar = 40 m/s (eastward).
And let the wind's velocity be Vwind = 3 m/s (southward).
To find the resultant velocity, we can subtract the wind's velocity vector from the car's velocity vector.
Resultant velocity = Vcar - Vwind
Since the velocities are in perpendicular directions, their components can be calculated separately.
The car's velocity can be broken down into eastward (x) and northward (y) components:
Vcar_x = 40 m/s
Vcar_y = 0 m/s
The wind's velocity can be broken down into eastward (x) and northward (y) components:
Vwind_x = 0 m/s
Vwind_y = -3 m/s
Now, let's subtract the wind's components from the car's components:
Resultant velocity in the x direction = Vcar_x - Vwind_x = 40 m/s - 0 m/s = 40 m/s (eastward)
Resultant velocity in the y direction = Vcar_y - Vwind_y = 0 m/s - (-3 m/s) = 3 m/s (northward)
To find the resultant speed, we can use the Pythagorean theorem:
Resultant speed = sqrt((Resultant velocity in the x direction)^2 + (Resultant velocity in the y direction)^2)
Resultant speed = sqrt((40 m/s)^2 + (3 m/s)^2)
Resultant speed ≈ sqrt(1600 m^2/s^2 + 9 m^2/s^2)
Resultant speed ≈ sqrt(1609 m^2/s^2)
Resultant speed ≈ 40.12 m/s (rounded to two decimal places)
To find the direction, we use trigonometry and inverse tangent (arctan):
Resultant direction = arctan(Resultant velocity in the y direction / Resultant velocity in the x direction)
Resultant direction = arctan(3 m/s / 40 m/s)
Resultant direction ≈ 4.3 degrees (rounded to one decimal place)
Therefore, the car's velocity relative to its original direction is approximately 40.12 m/s at an angle of 4.3 degrees north of east.