0
$\begingroup$

enter image description here

I found this question online and gave it a try, it seemed simple but I keep getting the wrong answer. One way or another my calculation keep showing me the answer B, which is regarded false. The answer key for this particular question is D.

So I'll breakdown what I did and please correct me if I'm wrong. Let us start. When the plane is travelling at a certain direction with a speed of 200km/h and affected by a wind from the West (85km/h), it's speed would be 217km/h (using the Pythagoras Theorem) and would be 23 degrees from the North. So basically the pilot needs to steer the plane 23 degrees due west of North to 'counteract' the wind. I just can't get how they got 25 degrees.

enter image description here

This was one of the solution diagram I found on the internet (and using sine inverse you'll get 25 degrees, and I thought for a while whether 200km/h was the speed with no wind or with, because it'll then makes sense, but looking back at the question it says 200km/h in 'still air'.

Can anyone help me here, any effort would be greatly appreciated.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ 'Speed in still air' equals speed $relative\ to\ air,$ whether or not the air is moving relative to the ground. So the diagram carries out the vector addition: "Velocity of plane relative to ground = velocity of plane relative to air + velocity of air relative to ground". $\endgroup$
    – Philip Wood
    May 10 '19 at 16:16
1
$\begingroup$

D is correct. In still air, means speed with respect to air, it is given as 200kmph. Now, let the airplane flies at $ \theta$° west of North. And the wind blows from West to East at 85 kmph . The resultant of the two gives the net velocity of plane in air. You have $\vec v_{plane|air}=\vec v_{plane}-\vec v_{air}$, so $\vec v_{plane}=\vec v_{plane|air}+\vec v_{air}$. The net velocity should be due north. So the horizontal component of $\vec v_{plane|air}$ balances the $\vec v_{air}$. You get:$200 \sin\theta=85$

Improve this answer
$\endgroup$
5
  • $\begingroup$ Yes. That is exactly true. The relative speed has to be same. The velocity of medium may add a component to the net velocity. The net velocity is measured with respect to stationery medium like ground $\endgroup$
    – Tojra
    May 11 '19 at 6:14
  • $\begingroup$ But when the plane is travelling at 200km/h and the wind 'pushes' the plane with a velocity of 85km/h, shouldn't the resultant velocity be 217 km/h (using pythagoras)? Or does the Velocity of the plane stays at 200km/h, no matter how much the velocity of the wind is? $\endgroup$
    – user226894
    May 11 '19 at 6:18
  • $\begingroup$ The problem is that for the net velocity to be due north, the velocity of plane wrt wind and the velocity of wind can't be perpendicular. You can't use Pythagoras to calculate the resultant for non-perpendicular vectors. In fact this is in the problem itself to calculate the angle. BTW, the magnitude of the resultant is not required to calculate the angle. The resultant would be $ 200cos\theta$ which is the vertical component of relative velocity of plane. The horizontal component gets cancelled by the velocity of wind so the resultant is due north. $\endgroup$
    – Tojra
    May 11 '19 at 6:25
  • $\begingroup$ Because if the Velocity of the plane and the wind were perpendicular the resultant would never be north (always at an angle) right? $\endgroup$
    – user226894
    May 11 '19 at 9:09
  • $\begingroup$ That is what I am explaining $\endgroup$
    – Tojra
    May 11 '19 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy