anonymous
  • anonymous
Hello! I have a question concerning X-T Graphs and finding instantaneous velocity. To determine the velocity of a curved line (representing changing motion), is it acceptable to draw any arbitrary tangent line that goes through that point?
Physics
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1450334279378:dw|
anonymous
  • anonymous
|dw:1450334426950:dw|
anonymous
  • anonymous
@ganeshie8

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anonymous
  • anonymous
Instantaneous velocity of that point *.
anonymous
  • anonymous
In order to determine the instantaneous velocity, we can analyze the slope of the position graph at a certain point. This value will yield that value that we're looking for here. Additionally, instantaneous velocity is mathematically defined as: \[\huge v_{inst}= \lim_{\Delta t \rightarrow 0}\frac{\Delta x}{\Delta t}\]
anonymous
  • anonymous
Also, we can extend the tangent line that touches the point of interest, as you have drawn. Let's take another example|dw:1450341059360:dw|
anonymous
  • anonymous
|dw:1450341179393:dw|
anonymous
  • anonymous
Here, \((t_0,d_0)\) is our particular point of interest. We can calculate the instantaneous velocity as:\[\huge \vec{v}_\text{inst}=\frac{x_2-x_1}{t_2-t_1}\]
anonymous
  • anonymous
@IrishBoy123 @Michele_Laino This should work out like this yea? It's been a while since I've had to analyze dist-time graphs for instantaneous values.
anonymous
  • anonymous
Oh and \(d\) is the same as \(x\). I unknowingly and mistakenly switched variables for position, but they're the same.
anonymous
  • anonymous
It also might be useful to note that it doesn't specifically matter where (t1,x1) or (t2,x2) are -- as long as they lie on the line of the slope. In other words, we could also use these two points to calculate the slope:|dw:1450341560229:dw|
anonymous
  • anonymous
Notice that these points aren't necessarily points on the graph -- they're points that touch the tangent line of our point of interest
Michele_Laino
  • Michele_Laino
for first part I think you are correct! Nevertheless, I think that the two points that we are considering, have to belong to the graph, and they have not to belong to the tangent line @CShrix
anonymous
  • anonymous
@Michele_Laino Why is that? Since we're analyzing the slope at that point, we can theoretically "expand" the tangent line and find the slope by looking at two points that lie on that tangent line
Michele_Laino
  • Michele_Laino
yes you are right! Nevertheless, the \(definition\) of derivative of a function, is the limit position of a secant line of the curve to which such tangent refers, so according to such definition, those points have to belong to such curve: |dw:1450450825090:dw|

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