Kinematics -- flying a plane in the wind to a destination

AI Thread Summary
A plane must adjust its flight angle to counteract a wind blowing at 92 km/h from 45° south of east while traveling at a speed of 730 km/h. The correct angle to fly is approximately 5.1° north of east, determined through vector addition and trigonometric calculations. Participants in the discussion emphasize the importance of accurately interpreting vector directions and using diagrams to visualize the problem. Miscalculations arise from incorrect assumptions about the relationships between angles and their sine values. Understanding these concepts is crucial for solving kinematic problems involving wind resistance.
Philly215
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Homework Statement


A plane is taking off from an airport directly west of the airport it wishes to touch down in. The plane in still air can travel with a constant speed of 730 km/h. If a wind is blowing constantly at 92 km/h [45° S of E], at what angle must the plane fly to compensate for the wind?

Homework Equations


Trig (sin, cos, tan) *no sin law*
V =d/t

The Attempt at a Solution


[/B]
730 km/h ÷ 92 km/h = 7.93 km/h
45/7.93= 5.67 ° [N of E]

* The answer for this question is 5.1° [N of E] and must be solved using resultant triangles, components and trig (no sin law). However, I am completely lost in solving this.
 
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Philly215 said:

Homework Statement


A plane is taking off from an airport directly west of the airport it wishes to touch down in. The plane in still air can travel with a constant speed of 730 km/h. If a wind is blowing constantly at 92 km/h [45° S of E], at what angle must the plane fly to compensate for the wind?

Homework Equations


Trig (sin, cos, tan) *no sin law*
V =d/t

The Attempt at a Solution


[/B]
730 km/h ÷ 92 km/h = 7.93 km/h
45/7.93= 5.67 ° [N of E]

* The answer for this question is 5.1° [N of E] and must be solved using resultant triangles, components and trig (no sin law). However, I am completely lost in solving this.
Welcome to the PF.

First, you should clarify what is meant by "[45° S of E]" -- does that mean the wind is coming from that direction, or blowing in that direction.

Next, draw a diagram showing the vectors of the plane's velocity (pointing left-to-right, angled either up or down), and the wind's velocity vector. When you place the vectors nose-to-tail (to add them), the resultant vector needs to point straight to the right (to the East).

Makes sense? Show us your sketch, please... :smile:
 
Philly215 said:
730 km/h ÷ 92 km/h = 7.93 km/h
You divided km/h by km/h and got a result with units of km/h? Shouldn't the units cancel?
45/7.93= 5.67 ° [N of E]
What was your thinking on the above step, dividing 45° by 7.93?

Did you draw a sketch of how the vectors should add? What direction should the resultant have?

Edit: Ah! berkeman got there ahead of me!
 
Your mistake comes from assumption that x is proportional to sin(x) for degrees and such high value as 45°, try plugging the exact value of sin(45°). sin(x)~x for small angle values, but for x in radians.
 
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