Finding Position from Equations of Motion at Different Times

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SUMMARY

The discussion focuses on finding the position from equations of motion at different times, specifically when given position at time ##t=t_0## instead of the standard ##t=0##. The primary equation referenced is $$x = x_0 + v_0t + \frac{1}{2} a t^2$$, where ##x_0## and ##v_0## are the initial position and velocity. The user proposes using the integral form $$\int_{x_0}^x dx' = \int_{t_0}^t v(t') dt'$$ to derive the position function. An alternative method involves substituting (t - t0) for t in the original equation, effectively shifting the time scale while maintaining the initial velocity at ##t=t_0##.

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PFuser1232
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Usually, equations of motion (with constant acceleration) are written in terms of values of position/velocity at time ##t=0##.
Take for example:

$$x = x_0 + v_0t + \frac{1}{2} a t^2$$

Where ##x_0## and ##v_0## are the values (at ##t=0##) of position and velocity respectively.

What if we're given position at some other time ##t=t_0##, instead of ##t=0##, and we're asked to find ##x## as a function of time?

What I do is I find a more general form of the equation I wrote above as follows:

$$\int_{x_0}^x dx' = \int_{t_0}^t v(t') dt'$$

Where ##x_0## now represents position at time ##t=t_0##.

Another way would be to eliminate the variable ##x(0)##

Is there any alternative approach?
 
Last edited:
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This approach is great. Of course, what you would find is that you could also just substitute (t - t0) for t in the original equation (i.e., shift the time scale), assuming that v0 is the velocity at t = t0.

Chet
 

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