Converting dt graph (and table) to velocity/time using two point

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To convert a distance/time graph into a velocity/time graph using the two-point method, one must understand that velocity is the rate of change of displacement over time, represented by the slope of the distance-time graph. A straight line on the distance-time graph indicates constant velocity, resulting in a horizontal line on the velocity-time graph. Conversely, a parabolic distance-time graph indicates changing velocity, leading to a sloped velocity-time graph. It is crucial to grasp the physical significance of these relationships rather than just memorizing equations. Understanding these concepts will facilitate the conversion process effectively.
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How can we convert a distance/time graph (that also contains a table of info) into velocity/time, using the two point method?
 
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celtics23 said:
How can we convert a distance/time graph (that also contains a table of info) into velocity/time, using the two point method?

What is the relationship between "velocity" and "distance"?

Zz.
 
d=vt
 
celtics23 said:
d=vt

'd' is displacement 've
you'll learn it afterward perhaps i see

i am giving you graph of speed vs time

if d-t graph is straight line at an angle α with x-axis then v-t graph is a line parallel to x axis
if d-t graph is parabolic then v-t graph is straight line at an angle θ with x axis
 
celtics23 said:
d=vt

Not quite.

You need to know that velocity is the time rate of change of displacement/position. In graph form, it corresponds to the slope at a given point in the displacement-time graph.

You should have enough here to answer the question.

Note that you need to know the physical meaning of all of this. It is not enough to just know/memorize the equation without understanding what it means.

Zz.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...

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