A small plane aims to fly directly east at 120. km/h, but a headwind is blowing 50. km/h at 22 degrees south of west, slowing the plane down. Find the plane's actual ground speed and direction.
Vr = 120km/h[0o] + 50km/h[202o,CCW].
Vr = 120+50*Cos202 + i50*sin202 =
120 - 46.4 - 18.7i = 73.6 - 18.7i, Q4.
Vr = 73.6 - 18.7i = 76[-14.2o] = 76km/h[14.2o] S. of E.
To find the plane's actual ground speed and direction, we need to use vector addition. Let's break down the given information into two vectors: the plane's velocity (eastward) and the wind's velocity (southwestward).
1. Plane's velocity: The plane aims to fly directly east at 120 km/h. This can be represented as a vector pointing eastward with a magnitude of 120 km/h.
2. Wind's velocity: The headwind is blowing at 50 km/h at 22 degrees south of west. To represent this as a vector, we need to find the components of the wind's velocity. The magnitude of the wind's velocity is 50 km/h, and we can use trigonometry to find its components:
- The south component (opposite side) is calculated as sin(22°) * 50 km/h.
- The west component (adjacent side) is calculated as cos(22°) * 50 km/h.
Now, let's calculate the components of the wind's velocity:
South component: sin(22°) * 50 km/h = 0.3827 * 50 km/h ≈ 19.14 km/h (rounded to two decimal places)
West component: cos(22°) * 50 km/h = 0.9240 * 50 km/h ≈ 46.20 km/h (rounded to two decimal places)
We have the components of the wind's velocity:
South component ≈ 19.14 km/h
West component ≈ 46.20 km/h
To find the actual ground speed and direction, we need to calculate the resultant vector of the plane's velocity (eastward) and the wind's velocity (southwestward).
To add vectors, we sum the corresponding components:
East component = Plane's velocity = 120 km/h
Southwest component = South component (wind's velocity) - West component (wind's velocity)
Southwest component ≈ 19.14 km/h - 46.20 km/h = -27.06 km/h (rounded to two decimal places)
Since the southwest component is negative, it means the resultant vector is in the opposite direction (northwestward).
Now, we can calculate the magnitude (ground speed) and direction of the resultant vector using the Pythagorean theorem and trigonometry:
Magnitude (ground speed) = √(East component^2 + Southwest component^2)
Magnitude ≈ √(120 km/h^2 + (-27.06 km/h)^2) ≈ √(14400 km^2/h^2 + 733.4436 km^2/h^2) ≈ √(15133.4436 km^2/h^2) ≈ 123 km/h (rounded to two decimal places)
Direction = atan(Southwest component / East component)
Direction ≈ atan(-27.06 km/h / 120 km/h) ≈ atan(-0.2255) ≈ -12.52° (rounded to two decimal places)
Therefore, the plane's actual ground speed is approximately 123 km/h, and its direction is approximately 12.52° south of west.