Relative Velocity Problem - find the angle

Click For Summary
SUMMARY

The discussion focuses on solving a relative velocity problem involving an airplane with a cruising speed of 547 mph and wind blowing at 84 mph at 40° north of east. The objective is to determine the heading direction for the plane to fly straight from Geordi to Data, located 982 miles at 28° west of north. Participants suggest using the Law of Sines to find the angle of the plane's heading, emphasizing the importance of vector addition to account for the wind's effect on the plane's trajectory.

PREREQUISITES
  • Understanding of vector addition in physics
  • Familiarity with the Law of Sines
  • Knowledge of trigonometric functions and angles
  • Basic concepts of relative velocity
NEXT STEPS
  • Study vector addition and its application in physics problems
  • Learn how to apply the Law of Sines in navigation scenarios
  • Explore the concept of resultant velocity in relative motion
  • Investigate online resources for solving relative velocity problems, such as HyperPhysics
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and vector analysis, as well as anyone interested in solving real-world navigation problems involving wind and aircraft movement.

jtw2e
Messages
27
Reaction score
0
Relative Velocity Problem -- find the angle

Homework Statement



An airplane has a cruising speed (speed of plane with respect to the air) of 547 mph. The wind is blowing at 84 mph 40° north of east. The plane needs to fly from Geordi to Data, which is 982 miles at 28° west of north.

Part A: What direction should the plane head to fly straight from Geordi to Data? (° W of N).

Part B: How long does it take for the plane to fly from Geordi to Data? (hours)

Homework Equations



Law of sines?

The Attempt at a Solution



http://www.flickr.com/photos/28413236@N03/5443668917/
5443668917_0148c5ab5e.jpg


Based on the figure I have made, which is probably incorrect if any pattern can be drawn from the previous 6 hours of failure:

I tried to use law of sines to find alpha:
84/sin(α) = 547/sin(50)

Using the inverse sine I get an angle of 6.76° for α. (This makes no sense as we're given that the direction from Geordi City to Data City is 28 degrees West of North. What is going on here?)
 
Last edited:
Physics news on Phys.org


I had a similar problem to this. I came across this website and it helped me through it.

http://hyperphysics.phy-astr.gsu.edu/hbase/airpw3.html#c2

You can click on "Details about calculation" and it will give a little more explanation. Then you can click "More detail on calculation" and it gives you another picture and a couple equations.

At first I wasn't sure where the equations came from. But if you sit there and think about what they are saying it makes sense.

Sorry I can't be more help, I've having trouble with the rest of my homework lol.
 


white_is said:
I had a similar problem to this. I came across this website and it helped me through it.

http://hyperphysics.phy-astr.gsu.edu/hbase/airpw3.html#c2

You can click on "Details about calculation" and it will give a little more explanation. Then you can click "More detail on calculation" and it gives you another picture and a couple equations.

At first I wasn't sure where the equations came from. But if you sit there and think about what they are saying it makes sense.

Sorry I can't be more help, I've having trouble with the rest of my homework lol.

Thanks, what they're saying does make sense. My problem still is illusive to me.
 


You know alpha, it is the direction of Data from Georgi with respect to North. The resultant velocity is the vector sum of the aeroplane velocity with respect to air, and the velocity of wind. You get a triangle with two sides known, and one angle also known (50+28) apply the Sine Law to get x, and from that beta, the direction of the plane with respect to North.

ehild
 

Attachments

  • aeroplane.JPG
    aeroplane.JPG
    15.9 KB · Views: 915


the plane will acquire the velocity of the wind too. wind would change its path. so just think how can you remove the effect of wind
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
25
Views
2K
Replies
14
Views
862
Replies
11
Views
3K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K