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Posted: 2/9/2014 6:34:15 AM EDT
Trying to figure out and visualize the technique behind calculating Instantaneous velocity (Vx = the limiting value of the ratio [delta x / delta t] as delta t approaches zero = dx/dt "the derivative of x with respect to t") If I am looking at a velocity time graph where a particle is moving along the x-axis and I know that from 0-3 seconds it has an average velocity of -2 m\s and from 3-6 seconds it has an average velocity of +2/3 m/s. How do I calculate the instantaneous velocity at any point in time (0-6sec)? Any help or recommendations would be appreciated, thanks.
Link Posted: 2/9/2014 7:10:11 AM EDT
[#1]
If the problem just gives you that information, you can't extrapolate a curve from that, unless there are other qualifiers like "the acceleration is constant". If you have an actual graph in front of you, you need to just read it.
Link Posted: 2/9/2014 8:37:03 AM EDT
[#2]
You noted that you are looking at a velocity vs. time graph. Did you mean position vs time? If you need velocity from a v vs. t graph, just find the time you want, and look for the velocity. That graph tells you what you want.

From the first few sentences, I think you meant a position vs time graph. If you want to calculate velocity from a position vs time graph, find the equations of the lines on the graph, and set them up as piecewise functions if there is more than one distinct section. Next, take the derivative of each section, giving you the velocity functions. Any given t can be plugged in to the resulting equation (or any t in the specified range, if your equation is piecewise), and you'll.have your instantaneous velocity.
Link Posted: 2/9/2014 8:53:09 AM EDT
[#3]
I was interpreting the graph wrong the graph itself is detailing the instantaneous velocity at the points in time out to 6 seconds. It gives an initial position of 1.0m @ 1.0s and it wanted me to find its position @ 6.0s. I was looking at the graph in terms of x&y not x&time lack of sleep is not a good thing, thanks brother
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