Vector Motion Problems: Finding Direction and Speed of a Plane

  • Thread starter Thread starter zack558
  • Start date Start date
AI Thread Summary
An airplane traveling at 265 km/h relative to the air needs to adjust its heading to counteract a 90 km/h wind blowing at 30° north of east to land due north. The wind's eastward and northward components are calculated as 77.9 km/h and 45 km/h, respectively. To determine the correct heading, vector addition is used, where the wind velocity vector and the airplane's airspeed vector combine to yield the ground velocity vector. The relationship between these vectors is expressed as the sum of the wind and airplane velocities equaling the ground velocity. Solving these equations will reveal both the necessary heading direction and the plane's speed relative to the ground.
zack558
Messages
1
Reaction score
0

Homework Statement



An airplane has a speed of 265 km/h relative to the air. There is a wind blowing at 90 km/h at 30° north of east relative to Earth. In which direction should the plane head to land at an airport due north of its present location(in degrees west of north)?
What is the plane's speed relative to the ground (km/h)?

Homework Equations



Vx0 = V0cos(O)
Vy0 = V0sin(O)


The Attempt at a Solution



VwindE = 90cos(30) = 77.9 km/h
VwindN = 90sin(30) = 45 km/h

I am trying to figure out what to do from here. It says that the plane's velocity is 265 km/h, but it doesn't say which way the plane is going. I am clueless on what to do to get the direction the plane should head. I know I don't have a lot of steps, but I'm completely stuck.
 
Physics news on Phys.org
Try to work out the problem, and show the steps.

One knows the air speed or speed of the airplane with respect to the air. One is trying to solve for the direction of the aircraft in the wind, in order to get to a destination. One knows the direction and speed of the wind.

Displacement vector = velocity vector * time.

Vector addition is part of the solution to this problem.
 
This might also help you with these kind of problems:

You have three vectors a, b, c:
The wind velocity a=(a1,a2)
The velocity of the airplane through the air b=(b1,b2)
The velocity of the airplane along the ground c=(c1,c2)

The relationship between these vectors is a+b=c, which is equivalent to the two equations
a1+b1=c1
a2+b2=c2

The solution to one of these equations gives the direction of the plane. This direction, along with the other equation gives the planes ground speed.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top