Nov 14, 2009 · Homework Statement i am having trouble converting an angular v vs. time to position vs. time graph. i know that the slope of the position is the velocity but the velocity graph has line going up and down. Homework Equations The Attempt at …

Dec 17, 2007 · Your first assertion, that acceleration occurs only at curves is correct. Velocity is only equal to zero when it (the position function) has zero slope (ie, no motion--- straight lines and relative max/mins) and acceleration equals zero when the velocity is constant (velocity is a straight line) and when the position graph switches concavity (inflection point).

Dec 9, 2016 · If I'm not mistaken at t = 0 the graph starts at half the amplitude or 5. Also the amplitude of this graph is 10, and at t = 0 angular velocity is also 0. 5 = 10*cos(0 + Φ) Φ = arccos(5/10) Φ = π/3 The answer is supposed to be -2π/3, I'm not sure what I did wrong or if I'm even doing the question right.

Sep 14, 2011 · Change of position per unit time is motion and hence, there is velocity there...for as long as velocity is the rate of change of position vs time, we can say that velocity is the derivative of position vs time curve. Acceleration is the rate of change of velocity per unit time...and so, acceleration is the derivative of the velocity vs time curve.

Mar 20, 2012 · You have a velocity vs time curve.You want the position vs time. Area under the curve,(this will be fairly simple to grasp) will be the value of position.(x=v*t) If the velocity curve is a straight line, the position is area of the triangle thus formed.

May 11, 2010 · Then I measured the distance of each mark made from the spark timer with a 2m stick and made a distance vs time graph from the data. I made a best fit polynomial line of degree 2 and found the equation of the fit line that turned out to have the form ___x^2+___x+___, where the ___ is where my coefficients are.

Mar 20, 2016 · This makes no sense. Between A and B, the position didn't change. Also, the units don't make sense -- you're multiplying a position (in m.) by a time (in sec.) so you wouldn't get meters as a result. In fact, you don't get anything meaningful by multiplying the …

Apr 11, 2007 · So, you have a graph of velocity vs. time. You integrate, you get position. What if you have a position vs time graph? For example, meters vs. seconds. If you integrate it, you will get meters seconds. Now, does this mean anything? For example, I've heard of "watt hours" but i simply don't understand how this makes any sense.

Sep 27, 2007 · What is the position-vs-time graph of a free-falling object (the simplest example for your F-vs-t graph)? If you are unsure, take a ball an throw it upwards, and study its motion. If that's too fast, try Galileo's experiment:

Sep 26, 2017 · But then I linearized the position vs Time graph (position vs Time^2.) However, the slope of the velocity vs Time graph i got from taking the slope of the tangents was about double of what I got from linearization. This doesn't make sense to me... Equations: Position vs Time: x = 9.396x^2-1.188x+6.014 Velocity vs Time: (from tangent slopes) 18.7798