After flying for 15 min in a wind blowing 42 km/h at an angle of 20° south of east, an airplane pilot is over a town that is 42 km due north of the starting point. What is the speed of the airplane relative to the air?

The ground velocity of the airplane is 168 km/hr in a northern direction. This equals the vector sum of the plane's velocity with respect to the air, PLUS the air speed with repect to the land.

Therefore you will need to subtract the air velocity from the ground velocity of the plane to get the plane's speed (and direction) with respect to the air.

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To determine the speed of the airplane relative to the air, we can break it down into components and use vector addition.

Let's start by considering the airplane's displacement relative to the air during the 15 minutes of flying. The airplane travels straight east, and the wind is blowing from a direction 20° south of east.

Since the airplane is flying east and the wind is coming from the south (20° south of east), we can calculate the horizontal component of the airplane's displacement relative to the air.

Horizontal component of airplane's displacement = (Speed of airplane relative to the air) * (time)

We are given that the airplane flew for 15 minutes. Converting 15 minutes to hours, we get:

Time = 15 minutes = 15/60 hours = 0.25 hours

Now, let's calculate the horizontal component of the airplane's displacement.

Horizontal component of airplane's displacement = (Speed of airplane relative to the air) * (time)

We'll denote the horizontal component as Dx and the speed of the airplane relative to the air as Va.

Dx = Va * 0.25

Next, we need to calculate the vertical component of the airplane's displacement relative to the air.

Since the wind is blowing southward, the vertical component of the airplane's displacement due to the wind will be in the south direction.

Let's denote the vertical component as Dy.

Dy = (Speed of wind) * (time)

We are given that the wind is blowing at a speed of 42 km/h, and the airplane flew for 15 minutes. Converting 15 minutes to hours, we get:

Time = 15 minutes = 15/60 hours = 0.25 hours

Dy = 42 km/h * 0.25

Now, let's calculate the net displacement of the airplane relative to the air.

Net displacement = sqrt(Dx^2 + Dy^2)

We are given that the town is 42 km due north of the starting point. So, the net displacement is equal to 42 km.

Therefore, we can set up the equation:

42 km = sqrt(Dx^2 + Dy^2)

Now you can solve this equation to find the value of Dx and Dy. Once you have the values of Dx and Dy, you can use them to find the speed of the airplane relative to the air using vector addition.