An airplane pilot wishes to fly due west. A wind of 82.0m/s is blowing toward the south.If the airspeed of the plane (its speed in still air) is 375.0 km/h, in which direction should the pilot head? What is the speed of the plane over the ground?

wind speed + plane speed= speed over ground

82S+375@direction=speed W

so,
375@direction= W-82S=W+82N

375SinTheta N+ 375CosThetw W is what you want.

82=375 sinTheta solve for Theta.
That is the direction N of W.
Speed over ground? 375CosTheta.

first convert km/h to m/s then squre both the numbers and then add them. after this you will have one big number. you then get the square root of this number and that is the speed of the plane over the ground

To determine in which direction the pilot should head, and the speed of the plane over the ground, we can use vector addition.

First, let's convert the given wind speed from m/s to km/h to match the airspeed of the plane.

82.0 m/s * 3600 s/h = 295,200 m/h
295,200 m/h * (1 km/1000 m) = 295.2 km/h

Now, we have the airspeed of the plane as 375.0 km/h and the wind speed as 295.2 km/h.

To find the direction the pilot should head, we need to calculate the resultant vector of the airspeed and the wind speed. We can achieve this by adding the two vectors together.

Considering the wind speed is blowing to the south, and the pilot wishes to fly west, we can assign the positive x-direction as west and the positive y-direction as south.

The vector for the airspeed of the plane can be represented as (375.0 km/h, 0°) since it is flying due west.

The vector for the wind speed can be represented as (0 km/h, -90°) since it is blowing toward the south.

To add these vectors, we can use vector addition techniques:

X-component (horizontal component):
375.0 km/h + 0 km/h = 375.0 km/h

Y-component (vertical component):
0 km/h + (-295.2 km/h) = -295.2 km/h

Now, we have the resultant vector in terms of its components: (375.0 km/h, -295.2 km/h)

To find the direction the pilot should head, we can use the inverse tangent function:

θ = tan^(-1)(y-component / x-component)
θ = tan^(-1)(-295.2 km/h / 375.0 km/h)

Calculating this using a calculator:
θ ≈ -36.5°

Therefore, the pilot should head in a direction of approximately 36.5° south of west.

To find the speed of the plane over the ground, we can use the Pythagorean theorem to calculate the magnitude of the resultant vector:

Speed over the ground = √((x-component)^2 + (y-component)^2)
Speed over the ground = √((375.0 km/h)^2 + (-295.2 km/h)^2)

Calculating this using a calculator:
Speed over the ground ≈ 477.6 km/h

Therefore, the speed of the plane over the ground is approximately 477.6 km/h.