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.

1. To find the distance and direction of the plane from the starting point, we can use vector addition.

Step 1: Convert the given distances and directions into x and y components.
- The first leg of the flight is 20 km in a direction 60 degrees north of east. We can break this down into x and y components.
x-component = 20 km * cos(60°) = 10 km
y-component = 20 km * sin(60°) = 17.32 km

- The second leg of the flight is 30 km straight east. Since it is in the east direction, the x-component will be 30 km and the y-component will be 0 km.

- The third leg of the flight is 10 km straight north. Since it is in the north direction, the x-component will be 0 km and the y-component will be 10 km.

Step 2: Add up the x and y components separately.
Total x-component = 10 km + 30 km + 0 km = 40 km
Total y-component = 17.32 km + 0 km + 10 km = 27.32 km

Step 3: Use the Pythagorean theorem to find the magnitude of the resultant vector (distance from the starting point).
Magnitude = sqrt((Total x-component)^2 + (Total y-component)^2)
Magnitude = sqrt((40 km)^2 + (27.32 km)^2) = 48.62 km

Step 4: Use trigonometry to find the direction of the resultant vector.
Direction = arctan(Total y-component / Total x-component)
Direction = arctan(27.32 km / 40 km) = 34.21 degrees

Therefore, the plane is approximately 48.62 km away from the starting point in a direction of 34.21 degrees north of east.

2. To find the resultant of the two given forces, we can use vector addition.

Step 1: Convert the given forces into x and y components.
- The first force of 100N at 170 degrees can be broken down into x and y components.
x-component = 100N * cos(170°)
y-component = 100N * sin(170°)

- The second force of 100N at 50 degrees can be broken down into x and y components.
x-component = 100N * cos(50°)
y-component = 100N * sin(50°)

Step 2: Add up the x and y components separately.
Total x-component = x-component of first force + x-component of second force
Total y-component = y-component of first force + y-component of second force

Step 3: Use the Pythagorean theorem to find the magnitude of the resultant vector.
Magnitude = sqrt((Total x-component)^2 + (Total y-component)^2)

Step 4: Use trigonometry to find the direction of the resultant vector.
Direction = arctan(Total y-component / Total x-component)

Therefore, by performing these calculations, we can find the magnitude and direction of the resultant force.

3. To find the direction and speed of the car relative to its original direction, we need to use vector subtraction.

Step 1: Convert the given magnitudes and directions into x and y components.
- The car is traveling eastwards at a speed of 40 m/s. Therefore, the x-component would be 40 m/s and the y-component would be 0 m/s.

- The wind is blowing southward at a speed of 3 m/s. Since it is in the south direction, the x-component would be 0 m/s and the y-component would be -3 m/s.

Step 2: Subtract the x and y components separately.
Total x-component = x-component of car - x-component of wind
Total y-component = y-component of car - y-component of wind

Step 3: Use the Pythagorean theorem to find the magnitude of the resultant vector (speed).
Speed = sqrt((Total x-component)^2 + (Total y-component)^2)

Step 4: Use trigonometry to find the direction of the resultant vector.
Direction = arctan(Total y-component / Total x-component)

By following these steps, we can determine the direction and speed of the car relative to its original direction.

1. To solve this problem, we can break down the distances traveled in each direction into their north and east components.

First, let's find the north and east components of the 20km distance traveled in a direction 60 degrees north of east:
North component = 20 km * sin(60 degrees) = 20 km * (√3/2) ≈ 17.32 km
East component = 20 km * cos(60 degrees) = 20 km * (1/2) = 10 km

Next, let's find the north and east components of the 30km distance traveled straight east:
North component = 30 km * sin(0 degrees) = 0 km
East component = 30 km * cos(0 degrees) = 30 km

Finally, let's find the north and east components of the 10km distance traveled straight north:
North component = 10 km * sin(90 degrees) = 10 km
East component = 10 km * cos(90 degrees) = 0 km

To find the final displacement of the plane, we need to sum up the north and east components separately:
North displacement = 17.32 km + 0 km + 10 km = 27.32 km
East displacement = 10 km + 30 km + 0 km = 40 km

Using these displacements, we can find the distance and direction of the plane from the starting point using the Pythagorean theorem and tangent function:
Distance = √(North displacement^2 + East displacement^2) = √(27.32 km^2 + 40 km^2) ≈ 48.00 km
Direction = arctan(North displacement / East displacement) = arctan(27.32 km / 40 km) ≈ 34.39 degrees north of east

Therefore, the plane is approximately 48.00 km away from the starting point, in a direction approximately 34.39 degrees north of east.

2. To find the resultant of two forces acting on a point object, we can use vector addition.

First, let's find the north and east components of the 100N force acting at 170 degrees:
North component = 100 N * sin(170 degrees) ≈ -98.5 N (negative because it points south)
East component = 100 N * cos(170 degrees) ≈ 34.42 N

Next, let's find the north and east components of the 100N force acting at 50 degrees:
North component = 100 N * sin(50 degrees) ≈ 76.60 N
East component = 100 N * cos(50 degrees) ≈ 64.28 N

To find the resultant of these forces, we add the north and east components separately:
North component of resultant = -98.5 N + 76.60 N ≈ -21.9 N (negative because it points south)
East component of resultant = 34.42 N + 64.28 N ≈ 98.70 N

Using these components, we can find the magnitude and direction of the resultant force using the Pythagorean theorem and the inverse tangent function:
Magnitude of resultant = √(North component^2 + East component^2) = √((-21.9 N)^2 + (98.70 N)^2) ≈ 101.02 N
Direction of resultant = arctan(North component / East component) = arctan((-21.9 N) / (98.70 N)) ≈ -12.1 degrees (negative because it points south)

Therefore, the resultant force is approximately 101.02 N, in a direction approximately 12.1 degrees south of east.

3. To find the direction and speed of the car relative to its original direction, we can use vector addition.

The car is traveling eastwards at a speed of 40 m/s, and the wind is blowing southward at a speed of 3 m/s.

First, let's find the north and east components of the 40 m/s velocity of the car:
North component = 40 m/s * sin(0 degrees) = 0 m/s
East component = 40 m/s * cos(0 degrees) = 40 m/s

Next, let's find the north and east components of the 3 m/s velocity of the wind:
North component = 3 m/s * sin(270 degrees) = -3 m/s (negative because it points south)
East component = 3 m/s * cos(270 degrees) = 0 m/s

To find the relative velocity of the car, we subtract the north and east components of the wind from the north and east components of the car's velocity:
North component of relative velocity = 0 m/s - (-3 m/s) = 3 m/s (north)
East component of relative velocity = 40 m/s - 0 m/s = 40 m/s (east)

Using these components, we can find the magnitude and direction of the relative velocity using the Pythagorean theorem and the inverse tangent function:
Magnitude of relative velocity = √(North component^2 + East component^2) = √((3 m/s)^2 + (40 m/s)^2) ≈ 40.25 m/s
Direction of relative velocity = arctan(North component / East component) = arctan((3 m/s) / (40 m/s)) ≈ 4.29 degrees (north of east)

Therefore, the car is traveling at a speed of approximately 40.25 m/s, in a direction approximately 4.29 degrees north of east relative to its original direction.

1. d = 20km @ 60deg. + 30km @ 0 deg. + 10km @ 90 deg.

X = hor. = 20cos60 + 30km + 0,
X = hor. = 10 + 30 = 40km.

Y = ver. = 20sin60 + 0 + 10sin90,
Y = ver. = 17.32 + 0 + 10 = 27.32km.

tanA = Y / X = 27.32 / 40 = 0.4330,
A = 23.4 Deg.

d = X / cosA = 40 / cos23.4 = 43.6km @
23.4 Deg. N of E.

2. R = 100N @ 170 Deg. + 100N @ 50 deg.
X = hor. = 100cos170 + 100cos50,
X = hor. = -98.48N + 64.28 = -34.20N.

Y = ver. = 100sin170 + 100sin50,
Y = ver. = 17.36 + 76.60N.

tanA = Y / X = 76.60 / -34.20 = -2.2399
A = -65.94 deg.,CW.
A = -65.94 + 180 = 114 deg.,CCW.

R = X / cosA = -34.20 / cos114 = 84.1N.
@ 114 deg.

3. Vc=40m/s @ 0 deg. + 3m/s @ 270 deg.

X = hor. = 40cos(0) + 3cos270,
X = hor. = 40 + 0 = 40m/s.
=
Y = ver. = 40sin(0) + 3sin270,
Y = ver. = 0 + (-3) = -3m/s.

tanA = Y / X = -3 / 40 = -0.075,
A = -4.29 deg.,CW. = 4.29 deg.,S of E.
A = -4.29 + 360 = 355.7 deg.,CCW.

Vc = X / cosA = 40 / cos355.7 = 40.1m/s
@ 355.7 deg.