What Is the Northern Component of an Airplane's Velocity?

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Homework Help Overview

The problem involves determining the northern component of an airplane's velocity, given that it travels at 146 km/h toward the northeast. The context is within the subject area of vector decomposition in physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of trigonometric functions, specifically sine and cosine, to resolve the velocity into its northward and eastward components. There is uncertainty about the application of the Pythagorean theorem and the correct interpretation of the angle associated with northeast direction.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications on the mathematical approach. There is acknowledgment of the need to use sine and cosine for calculating the components, and some participants are correcting terminology and notation. However, no explicit consensus on the final calculations has been reached.

Contextual Notes

There is a mention of potential confusion regarding the notation of speed (km=h vs. km/h) and the assumption that northeast corresponds to a 45-degree angle. Participants are also reflecting on the accuracy of their mathematical computations.

anglum
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Components of velocity

Homework Statement



An airplane travels at 146 km=h toward the
northeast.
What is the northern component of its ve-
locity? Answer in units of km=h.

Homework Equations



asquared + bsquared = csquared

The Attempt at a Solution



however i am not sure if u use the pythagorean formula to solve this?
 
Last edited:
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Yes, you're on the right track. (Assuming by "North East" they mean that 45 degree angle between north and east. That means NorthEast would be the hypotenuse/resultant of a two vectors: 1 north and 1 east of equal magnitudes.)
 
First, the plane speed is 146 km/h (not km=h).

Next, you find the components in the north and east directions by multiplying the hypoteneuse by the sine or cosine of the appropriate angles...
 
ooo i have to do sin/cosine of the angle
 
so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h
 
Last edited:
anglum said:
so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h

Yes. That's an unusual way to write it, however. More like this (I'll use latex):

v_y = 146 km/h * sin(45)
 
anglum said:
so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h

Well, except for the math you did. The sin and cos of 45 degrees should be the same...
 
ok thanks guys i got it i was doing some bad math... thank you
 
Last edited:

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