Finding the magnitude of velocity given vx and vy

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    Magnitude Velocity
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Homework Help Overview

The discussion centers around determining the magnitude of velocity given the horizontal and vertical components, vx and vy, in the context of a lab assignment. The original poster expresses uncertainty about how to derive the equation for velocity without a calculus background.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the components of velocity and the use of vector addition, referencing Pythagorean theorem to relate vx and vy to the resultant velocity v.

Discussion Status

Some participants have provided insights into vector addition and the geometric interpretation of the problem. The original poster acknowledges a misunderstanding and indicates a readiness to apply the concepts discussed.

Contextual Notes

The original poster mentions that the course has not covered calculus, which may influence their approach to the problem. There is an indication that vector addition has been covered in class, which may guide their understanding.

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Homework Statement


I am working on a lab, and I have come up with graphs (and the data set) for vy vs. time and vx vs. time, but I'm not sure how to determine v? I just need to come up with the equation so solve for v, but we haven't done anything calc oriented in the course yet.


Homework Equations





The Attempt at a Solution

 
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It has nothing to do with calculus, just simple vector addition. The horizontal component of v is vx while the vertical component of v is vy. Hence, by Pythagora's theorem, we have [tex]v^{2} = v_{x}^{2} + v_{y}^{2}[/tex]
 
Have you seen vectors? There is a right triangle with vx as one right side and vy as the other. You are looking for the speed v which is the hypotenuse of the triangle. So ...
 
Okay thank you both for your help.
I guess I just misunderstood the question, we have covered adding vectors in the course so I am able to do that.

thanks again!
 

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