An airplane is flying with a heading of 60 degrees east of north at 450 mph. A wind is blowing at 37 mph from the north. Find the plane's ground speed to the nearest mph and direction to the nearest degree.

Thanks.

the plane is traveling South with the wind at 37 miles per hour

It is flying North in the air at 450 cos 60 = 225
so its speed North = 225 - 37 = 188
its speed East is 450 sin 60 = 389
so
total speed = sqrt(188^2 + 389^2)
=sqrt (186665)
= 432 mph
tan angle east of north = 389/188 = 2.07
angle = 64.2 = 64 degrees

Why did the airplane become friends with the wind? Because they were both heading in the same direction - towards north! Alright, let's calculate the plane's ground speed and direction.

To find the ground speed, we need to consider the effect of the wind on the plane's speed. Since the wind is blowing from the north, it will affect the plane's speed towards the east. So, we need to find the horizontal component of the plane's velocity.

To do that, we break down the plane's velocity into its northward and eastward components. The northward component remains unaffected by the wind because the wind is blowing perpendicular to it. However, the eastward component is affected by both the plane's velocity and the wind's velocity.

The northward component of the plane's velocity is given by:
Vn = Velocity * sin(angle)
Vn = 450 mph * sin(60°)
Vn ≈ 389.71 mph.

The eastward component of the plane's velocity is given by:
Ve = Velocity * cos(angle)
Ve = 450 mph * cos(60°)
Ve ≈ 225 mph.

Now, we can calculate the ground speed by adding the horizontal components of the plane's velocity and the wind's velocity:
Vg = sqrt(Vn^2 + Ve^2)
Vg = sqrt((389.71 mph)^2 + (37 mph)^2)
Vg ≈ 391.41 mph (to the nearest mph).

The direction of the ground speed can be found using trigonometry as well. The angle can be calculated as:
angle = arctan(Ve / Vn)
angle = arctan(225 mph / 389.71 mph)
angle ≈ 30.59° (to the nearest degree).

So, the plane's ground speed is approximately 391 mph, and its direction is approximately 31° east of north.

To find the plane's ground speed, we can use vector addition.

Let's break down the velocities into their components.

The velocity of the plane can be broken down into two components: a north component and an east component.

The north component of the plane's velocity is given by:
v_n = 450 mph * cos(60 degrees)

The east component of the plane's velocity is given by:
v_e = 450 mph * sin(60 degrees)

The velocity of the wind is purely northward, so its north component is 37 mph and its east component is 0 mph.

Now let's add the components of the plane's velocity and the wind's velocity to get the resultant velocity.

The north component of the resultant velocity is:
v_n_resultant = v_n + wind_n
v_n_resultant = v_n + 37 mph

The east component of the resultant velocity is:
v_e_resultant = v_e + wind_e
v_e_resultant = v_e + 0 mph

To find the magnitude of the resultant velocity (the ground speed), we can use the Pythagorean theorem:
ground_speed = sqrt((v_n_resultant)^2 + (v_e_resultant)^2)

To find the direction of the resultant velocity, we can use the inverse tangent function:
direction = atan2(v_e_resultant, v_n_resultant)

Now, let's calculate the values:

v_n = 450 mph * cos(60 degrees)
v_n = 450 mph * 0.5
v_n = 225 mph

v_e = 450 mph * sin(60 degrees)
v_e = 450 mph * √3/2
v_e ≈ 389.7 mph

v_n_resultant = v_n + 37 mph
v_n_resultant = 225 mph + 37 mph
v_n_resultant = 262 mph

v_e_resultant = v_e + 0 mph
v_e_resultant = 389.7 mph + 0 mph
v_e_resultant = 389.7 mph

ground_speed = sqrt((v_n_resultant)^2 + (v_e_resultant)^2)
ground_speed ≈ sqrt((262 mph)^2 + (389.7 mph)^2)
ground_speed ≈ sqrt(68644 + 151898.09)
ground_speed ≈ sqrt(220542.09)
ground_speed ≈ 469.63 mph (rounded to the nearest mph)

direction = atan2(v_e_resultant, v_n_resultant)
direction = atan2(389.7 mph, 262 mph)
direction ≈ 53.78 degrees (rounded to the nearest degree)

Therefore, the plane's ground speed is approximately 470 mph in a direction of 54 degrees east of north.

To find the plane's ground speed and direction, we can first break down the velocity vectors into their components.

The airplane's velocity vector can be split into two components: one in the north direction and one in the east direction. We can use trigonometry to find these components.

The northward component of the airplane's velocity can be found by multiplying the magnitude of the velocity (450 mph) by the cosine of the angle east of north (60 degrees).

Northward component = 450 mph * cos(60 degrees) = 450 mph * 0.5 = 225 mph

The eastward component of the airplane's velocity can be found by multiplying the magnitude of the velocity (450 mph) by the sine of the angle east of north (60 degrees).

Eastward component = 450 mph * sin(60 degrees) = 450 mph * 0.866 = 389.7 mph

Now let's consider the wind vector. Since the wind is only in the north direction, its northward component will be equal to its magnitude (37 mph) and the eastward component will be zero.

Northward component of wind = 37 mph
Eastward component of wind = 0 mph

To find the resultant velocity (ground speed), we can add the components of the airplane's velocity and the wind velocity:

Northward component of resultant velocity = northward component of airplane's velocity + northward component of wind
= 225 mph + 37 mph = 262 mph

Eastward component of resultant velocity = eastward component of airplane's velocity + eastward component of wind
= 389.7 mph + 0 mph = 389.7 mph

The ground speed is the magnitude of the resultant velocity vector:

Ground speed = √[(Northward component of resultant velocity)^2 + (Eastward component of resultant velocity)^2]
= √[(262 mph)^2 + (389.7 mph)^2] ≈ √(68644 + 151804) ≈ √220448 ≈ 469.6 mph ≈ 470 mph (rounded to the nearest mph)

The direction of the resultant velocity can be found using trigonometry as well:

Direction = arctan(Eastward component of resultant velocity / Northward component of resultant velocity)
= arctan(389.7 mph / 262 mph) ≈ arctan(1.487) ≈ 56.7 degrees (rounded to the nearest degree)

Therefore, the plane's ground speed is approximately 470 mph, and its direction is approximately 57 degrees east of north.