The pilot of an aircraft wishes to fly due
west in a 50.0 km/h wind blowing toward
the south. The speed of the aircraft in the absence of a wind is 228 km/h.
a) How many degrees from west should the
aircraft head? Let clockwise be positive.
Answer in units of degrees.
b) What should the plane’s speed be relative to the ground? Answer in units of km/h
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I drew this a free body diagram of this
You should get a triangle when you do this
where 228 is your hypothenuse is 228
so the formula to to find the missing side is
c^2=a^2+c^2
where c=228
and a=50
so the missing side b = 222.45
So using the triogonometry
cos^-1(222.45/228) = 12.667 degrees
and speed relative to the ground is the missing side
222.45 km/h
To solve this problem, we can break it down into two components: direction and speed.
a) Finding the direction:
Since the wind is blowing from the south and the pilot wants to fly due west, we need to calculate the angle between west and the direction the plane should head. We can use trigonometry to do this.
Let's define angle θ as the angle the plane should head relative to west. By drawing a triangle with the wind as the hypotenuse, we can determine that the opposite side (side opposite to the angle θ) represents the south component of the wind, which is the same as the component the plane needs to counteract.
Using trigonometry, we can use the tangent function:
tan(θ) = opposite/adjacent
tan(θ) = South velocity (50.0 km/h) / Plane's airspeed (228 km/h)
Solving for θ:
θ= arctan(South velocity / Plane's airspeed)
θ= arctan(50.0 km/h / 228 km/h)
Using a calculator, we find that θ is approximately 12.52 degrees. Since we defined clockwise as positive, the plane should head 12.52 degrees south of due west.
b) Finding the speed relative to the ground:
The speed of the plane relative to the ground can be calculated using vector addition. We need to find the resultant vector of the plane's airspeed and the wind velocity.
Let's define the ground speed as GS, the airspeed as AS, and the wind velocity as WV. The ground speed is the magnitude of the resultant vector.
Using the Pythagorean Theorem:
GS^2 = AS^2 + WV^2
Plugging in the values:
GS^2 = (228 km/h)^2 + (50.0 km/h)^2
Solving for GS:
GS = √[(228 km/h)^2 + (50.0 km/h)^2]
Using a calculator, we find that GS is approximately 235.13 km/h. So the plane's speed relative to the ground is approximately 235.13 km/h.
To summarize:
a) The pilot should head approximately 12.52 degrees south of due west.
b) The plane's speed relative to the ground is approximately 235.13 km/h.