I have a question about an airspeed/groundspeed word problem

In summary, the plane is heading south and climbing at a rate of 60 km/hr. If the airspeed is 540 km/hr, what is the ground speed of the plane? The Attempt at a Solution:To solve for the ground speed of the plane, you need to combine the airspeed (540 km/hr) with the direction of the flight (due south). To do this, you need to use the Pythagorean theorem. First, you need to find the x component of the airspeed. This is done by multiplying the airspeed by the cosine of the direction of the flight (in this case, due south). Next, you need to find
  • #1
PatrickE
6
0

Homework Statement



A plane is heading due south and climbing at the rate of 60 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 110 km/hr to the northwest, what is the ground speed of the plane?

Homework Equations



Pythagorean theorem

The Attempt at a Solution



√(540^2+110^2). I tried adding different vectors using this method, but none of the results are correct.
 
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  • #2
South and northwest do not make a right angle. you need to resolve each motion into suitable orthogonal components and add the components separately. Don't forget that some of the airspeed is upwards, and this won't count at all towards groundspeed.
 
  • #3
Thanks for the reply. Now what?

I'm still confused about exactly how I break down the components. Am I at least using the right formula? Please no vague answers.
 
  • #4
Well, if there are no other respondents:

It says the plane is "heading" south. In aircraft parlance that would mean that it's pointed south but not necessarily flying due south.

I see the problem solvable only if we assune the direction of the flight is due south.

So: let x = east, y = north, z = up. What is the x component of indicated airspeed IAS, given that the plane's x ground speed is v_px? Recall there is a wind blowing to the northwest which his IAS indicator will register.
and same question for indicated y airspeed if its y ground speed is v_py?
The how about combining those velocity components along with v_pz with the total airspeed to solve for v_px and v_py?

(Hint: what is v_px right off the bat?)
 
  • #5
I don't know what the x component of the airspeed is. How would I find it?
 
  • #6
PatrickE said:
I don't know what the x component of the airspeed is. How would I find it?

It's going to be the vector sum of the plane's airspeed without any wind, plus you have to add the x component of wind, right?
 
  • #7
You mean the vector sum of 540 km/hr South? Since there are no degree measures in the problem, should I assume that it means exactly Northwest at 45 degrees?
 
  • #8
It would help if I had any idea how to draw this scenario.
 
  • #9
rude man said:
It says the plane is "heading" south. In aircraft parlance that would mean that it's pointed south but not necessarily flying due south.

I see the problem solvable only if we assune the direction of the flight is due south.
I agree that south is the direction relative to the wind. Not sure if that's what you meant by the last sentence above. But it's solvable with either interpretation.
The wind velocity is one vector, the velocity of the plane relative to the wind another. Just add the two (but discard the vertical component).
PatrickE: I don't know what the x component of the airspeed is. How would I find it?
X component of the windspeed. Airspeed means the speed of the plane relative to the air. If something is moving northwest at 1kph, how much further west is it after one hour?
 
  • #10
haruspex said:
I agree that south is the direction relative to the wind. Not sure if that's what you meant by the last sentence above. But it's solvable with either interpretation.

'Heading' is not direction wrt a wind direction. It's the direction of flight plus/minus the crab angle. The crab angle varies with aircraft aerodynamical properties, speed, etc.
 
  • #11
If you go northwest for an hour at 1 kph you should go about 0.7 km west, right?
 
  • #12
PatrickE said:
If you go northwest for an hour at 1 kph you should go about 0.7 km west, right?
Yes, or more precisely 1/√2 km. so if The wind is blowing at 110kph NW, what components does that velocity have in the north and west directions?
 
  • #13
rude man said:
'Heading' is not direction wrt a wind direction.

I stand corrected. But do we agree that that's probably how it's intended in the questioin?
 
  • #14
PatrickE said:
If you go northwest for an hour at 1 kph you should go about 0.7 km west, right?

Well, you're not really 'going' NW. That's the direction of the wind. You're 'going' south. But your IAS indicator registers wind as though it were the airplane flying; it can't tell the difference between wind and aircraft motion. They're both 'wind' to it.

For the IAS indicator, yes, if it's going south and the wind is nw then it will register +0.707 times the actual windspeed of 110 km/h, pplus the IAS due to the plane's own motion.

What you're trying to do is combine wind and aircraft motion to determine the aircraft ground speed.
 
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  • #15
haruspex said:
I stand corrected. But do we agree that that's probably how it's intended in the questioin?

I don't know. I am assuming the plane continues on a due-south path. Like I said - could be wrong!

I do know this - an aircraft is not subject to wind motion the way a canoe is to moving water. So you can't just add wind and aircraft velocities the way you can with a boat and a moving river. In other words, there is no 'slippage' assumed with a boat but there is with a plane.
 
  • #16
rude man said:
Well, you're not really 'going' NW. That's the direction of the wind. You're 'going' south.
I was just using that to get PatrickE to figure out how to resolve the windspeed into components.
I don't know. I am assuming the plane continues on a due-south path. Like I said - could be wrong!

I do know this - an aircraft is not subject to wind motion the way a canoe is to moving water. So you can't just add wind and aircraft velocities the way you can with a boat and a moving river. In other words, there is no 'slippage' assumed with a boat but there is with a plane.
As I understand it, an airspeed indicator measures just that - the speed (and direction) of the plane relative to the air. I don't see how it could do anything else. Note that this is nothing to do with how the plane creates that motion. It could be a rocket; could be a glider pulled along by cable by a truck on the ground.
 
  • #17
haruspex said:
I was just using that to get PatrickE to figure out how to resolve the windspeed into components.

As I understand it, an airspeed indicator measures just that - the speed (and direction) of the plane relative to the air. .

Speed, yes. Direction, no.

It measures pitot vs. static pressure and determines the indicated airspeed from that.
 

Related to I have a question about an airspeed/groundspeed word problem

1. What is the difference between airspeed and groundspeed?

Airspeed is the speed at which an aircraft is moving through the air. It is measured by an airspeed indicator and is affected by factors such as wind and air density. Groundspeed, on the other hand, is the speed at which an aircraft is moving over the ground. It takes into account the aircraft's airspeed as well as the effects of wind and can be calculated using a combination of airspeed and wind direction/speed.

2. How do you solve an airspeed/groundspeed word problem?

The first step in solving an airspeed/groundspeed word problem is to identify all the given information, such as the aircraft's airspeed, wind direction and speed, and distance traveled. Then, use the appropriate formula to calculate either the airspeed or groundspeed. Finally, check your answer and make sure it makes sense in the context of the problem.

3. What formula should I use to calculate airspeed or groundspeed?

The formula for calculating airspeed is Airspeed = Groundspeed + Headwind or Tailwind. For calculating groundspeed, the formula is Groundspeed = Airspeed - Headwind or Tailwind. In both cases, the direction of the wind (headwind or tailwind) is important and must be factored into the calculation.

4. What units are typically used for airspeed and groundspeed?

Airspeed and groundspeed are usually measured in knots (nautical miles per hour) or miles per hour (mph). However, other units such as kilometers per hour (km/h) or meters per second (m/s) may also be used. It is important to check the units given in the problem and convert them if necessary to ensure accurate calculations.

5. Are there any common mistakes to avoid when solving airspeed/groundspeed word problems?

One common mistake is forgetting to factor in the direction of the wind. It is important to pay attention to whether the wind is a headwind or a tailwind and adjust the formula accordingly. Another mistake is using the wrong units, so make sure to double-check and convert if necessary. Lastly, be sure to carefully read the problem and understand what is being asked before attempting to solve it.

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