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### #ActualJ.B47

Posted 01 October 2012 - 01:02 PM

Actually, this is basically the same in 2D and 3D and it all follows directly from the definition of velocity.

velocity = distance / time, time = distance / velocity. All you need is the actual distance, which can be computed using vector subtraction (x2-x, y2-y) using the Pythagorean theorem: distance = sqrt(a^2+b^2). For 3D, you add the third dimension and are done.

Hi rnlf. Thanks for pointing me up! I have another question, how do i get X2, Y2 position depending of the given time, velocity, and distance? Do i need the slope of the line?
http://en.wikipedia.org/wiki/Slope

### #2J.B47

Posted 01 October 2012 - 01:01 PM

Actually, this is basically the same in 2D and 3D and it all follows directly from the definition of velocity.

velocity = distance / time, time = distance / velocity. All you need is the actual distance, which can be computed using vector subtraction (x2-x, y2-y) using the Pythagorean theorem: distance = sqrt(a^2+b^2). For 3D, you add the third dimension and are done.

Hi rnlf. Thanks for pointing me up! I have another question, how do i get X2, Y2 position depending of the given time, velocity, and distance? Do i need the slope of the rect?
http://en.wikipedia.org/wiki/Slope

### #1J.B47

Posted 01 October 2012 - 11:39 AM

Actually, this is basically the same in 2D and 3D and it all follows directly from the definition of velocity.

velocity = distance / time, time = distance / velocity. All you need is the actual distance, which can be computed using vector subtraction (x2-x, y2-y) using the Pythagorean theorem: distance = sqrt(a^2+b^2). For 3D, you add the third dimension and are done.

Hi rnlf. Thanks for pointing me up! I have another question, how do i get X2, Y2 position depending of the given time, velocity, and distance?

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