Vectors componenets relative motion help

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SUMMARY

The discussion focuses on calculating the relative velocity of two airplanes, Plane 1 and Plane 2, with speeds of 12.1 m/s due north and 6.3 m/s at 21.2° north of west, respectively. The correct approach to find the velocity of Plane 1 relative to Plane 2 involves vector subtraction rather than addition. The resultant velocity magnitude is calculated as 15.53 m/s at an angle of 67.78° north of east. The confusion arises in determining the direction of Plane 2's velocity relative to Plane 1, which requires understanding the vector subtraction method.

PREREQUISITES
  • Understanding of vector components and their graphical representation
  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Familiarity with relative velocity concepts in physics
  • Ability to perform vector addition and subtraction
NEXT STEPS
  • Study vector subtraction techniques in relative motion problems
  • Learn how to resolve vectors into components using trigonometric functions
  • Explore the concept of relative velocity in different frames of reference
  • Practice problems involving multiple vectors and their resultant magnitudes and directions
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators seeking to clarify concepts of relative motion and vector calculations.

P944
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Homework Statement


Hello. I was wondering if you guys could help me tell me why i am getting this problem wrong.

Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 12.1 m/s due north. Plane 2 taxies with a speed of 6.3 m/s in a direction 21.2° north of west.
(a) What are the direction and magnitude of the velocity of plane 1 relative to plane 2?
Direction _______° north of east
Magnitude ________m/s

(b) What are the direction and magnitude of the velocity of plane 2 relative to plane 1?
Direction ______° south of west
Magnitude ________ m/s

Homework Equations






The Attempt at a Solution



I First drew out a parallelogram with the magnitude and direction of plane 1 then drew plane 2's tail to the head of plane 1 to visualize the problem. Next i used components of a vector to solve for the magnitude and direction.

plane 1 -> x = 12.1 cos 90 =0
plane 2 -> x = 6.3 cos 21.2 = 5.874

plane 1 -> y = 12.1 sin 90 = 12.1
plane 2 -> y = 6.3 sin 21.2 = 2.278

I added the vectors (5.874 m/s) x + (14.38 m/s) y

To find the magnitude of the resultant i got sq root (5.874 m/s)^2 + (14.38 m/s)^2 = 15.53

theta = inverse tan (14.378/5.8736) = 67.78 degrees

b) I was confused on this one. I thought that from plane 2 the angle would be 67.78 + 180 = 247.8 degrees with the same magnitude of 15.53?

I would greatly appreciate if someone could help me out and let me know what i did wrong. Thanks so much! :)
 
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Hi P944! :smile:
P944 said:
Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 12.1 m/s due north. Plane 2 taxies with a speed of 6.3 m/s in a direction 21.2° north of west.
(a) What are the direction and magnitude of the velocity of plane 1 relative to plane 2?

I First drew out a parallelogram with the magnitude and direction of plane 1 then drew plane 2's tail to the head of plane 1

No, that will add the velocities.

For a relative velocity, you always need to subtract.

If g is "ground", then you want V12, which = V1g + Vg2 = V1g - V2g :wink:

(if you draw arrows, this should be obvious)
 

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