An airplane whose airspeed is 580 km/h is supposed to fly a straight line 38 degrees north of east. a steady 82 km/h wind is blowing from the north. in what direction should the plane head?

44.4° north of east

Explanation:

The plane's airspeed (speed relative to the air) is 580 km/h.

The wind speed is 82 km/h from the north (towards the south).

We want the plane's velocity relative to the ground to be at an angle of 38 degrees north of east.

The plane's velocity relative to the ground is the sum of the velocity relative to the air and the velocity of the wind.  (See attached diagram.)

v = va + vw

In the x direction:

vₓ = vaₓ + vwₓ

v cos 38 = 580 cos θ + 0

In the y direction:

vₓ = vaₓ + vwₓᵧ

v sin 38 = 580 sin θ − 82

Solve the system of equations.  Start by solving for v in the first equation and substituting into the second:

v = 580 cos θ / cos 38

(580 cos θ / cos 38) sin 38 = 580 sin θ − 82

580 cos θ tan 38 = 580 sin θ − 82

Problem here is we have both sin θ and cos θ.  But if we square both sides...

(580 cos θ tan 38)² = (580 sin θ − 82)²

(580 tan 38)² cos² θ = 580² sin² θ − 2(580 sin θ)(82) + 82²

(580 tan 38)² cos² θ = 580² sin² θ − 95120 sin θ + 82²

Using Pythagorean identity:

(580 tan 38)² (1 − sin² θ) = 580² sin² θ − 95120 sin θ + 82²

(580 tan 38)² − (580 tan 38)² sin² θ = 580² sin² θ − 95120 sin θ + 82²

0 = [ (580 tan 38)² + 580² ] sin² θ − 95120 sin θ + 82² − (580 tan 38)²

0 ≈ 541741 sin² θ − 95120 sin θ − 198617

Now we can solve with quadratic formula:

sin θ ≈ [ 95120 ± √(95120² - 4(541741)(-198617)) ] / 2(541741)

sin θ ≈ (95120 ± 662905.5) / 1083482

sin θ ≈ -0.5240, 0.6996

Since we know θ is in the first quadrant, we can ignore the negative answer.

sin θ ≈ 0.6996

θ ≈ 44.4°

The plane should fly 44.4° north of east in order to stay on the desired path.

d. platinum would have the greatest temperature increase because it has the lowest specific heat.

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