Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 12 m/s due north. Plane 2 taxies with a speed of 8.5 m/s in a direction 20 degrees north of west.
Incomplete.
To find the resultant velocity of the two airplanes, we need to add the individual velocities as vectors.
First, let's break down the velocities of the two airplanes into their horizontal and vertical components.
For Plane 1:
Velocity (magnitude and direction) = 12 m/s due north
Horizontal component = 0 m/s (since it's moving directly north)
Vertical component = 12 m/s
For Plane 2:
Velocity (magnitude and direction) = 8.5 m/s at 20 degrees north of west
Horizontal component = 8.5 m/s * cos(20°)
Vertical component = -8.5 m/s * sin(20°) (negative sign because it's moving opposite to north direction)
Now, let's add the horizontal and vertical components separately.
Horizontal component: 8.5 m/s * cos(20°) = 8.119 m/s (towards east)
Vertical component: 12 m/s - 8.5 m/s * sin(20°) = 2.455 m/s (towards north)
Now, we have the horizontal and vertical components of the resultant velocity of the two airplanes.
To find the magnitude and direction of the resultant velocity, we can use the Pythagorean theorem and the tangent function.
Magnitude of the resultant velocity:
Resultant velocity = √[(Horizontal component)^2 + (Vertical component)^2]
= √[(8.119 m/s)^2 + (2.455 m/s)^2]
≈ √(66.06 + 6.025)
≈ √72.085
≈ 8.49 m/s
Direction of the resultant velocity:
Direction = tan^(-1)(Vertical component / Horizontal component)
= tan^(-1)(2.455 m/s / 8.119 m/s)
≈ tan^(-1)(0.3017)
≈ 17.9°
Therefore, the resultant velocity of the two airplanes is approximately 8.49 m/s at 17.9 degrees north of east.