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Derive a formula for motion with constant acceleration and constant deceleration

  1. Jul 11, 2012 #1

    LoA

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    1. The problem statement, all variables and given/known data

    A subway train travels over a distance [itex] s [/itex] in [itex] t [/itex] seconds. it starts from rest and ends at rest. In the first part of its journey it moves with constant acceleration [itex] f [/itex] and in the second part with constant deceleration [itex] r \, [/itex].

    Show that [itex] s \, = \, \frac {[\frac {fr} {f \, + \, r}] \, t^2} {2} [/itex]

    2. Relevant equations

    I know that [itex] s \,= \, (\frac {1}{2})(-r )t^2\, +\, v_it[/itex], where [itex] v_i \,=\, ft[/itex] but I'm not sure where to go from there. In particular, I can't figure out how to connect the seemingly separate equations for distance generated by the different accelerations into one function of time for the entire interval.

    3. The attempt at a solution

    By assuming that total acceleration is a sum of the given accelerations, I've gotten something that looks awfully close to the desired result, but am still not quite there:
    [itex] s \,=\,\frac{1}{2} \,(f\,+\,r)\,t^2 \, + \,v_i\,t [/itex]

    I feel like I'm missing something.
     
    Last edited: Jul 11, 2012
  2. jcsd
  3. Jul 11, 2012 #2
    You have to integrate the aceleration two times to get the movement ecuation.

    You know that:

    [itex]x (t) = x_0 + v_0 (t-t_0) + \frac{1}{2} a (t-t_0)^2[/itex]

    [itex]v (t) = v_0+ a (t-t_0)[/itex]

    Part one we start from time, position and initial velocity cero and also positive aceleration f:

    [itex]x (t) =\frac{1}{2} f t^2[/itex]

    [itex]v (t) = a t[/itex]

    We know that at a certain time [itex]t_1[/itex] we reach a position [itex]x_1[/itex] with a velocity [itex]v_1[/itex] and from that point the movil starts desacelerating with aceleration negative r.

    [itex]x (t) = x_0 + v_0 (t-t_0) - \frac{1}{2} a (t-t_0)^2[/itex]

    [itex]v (t) = v_0 - a (t-t_0)[/itex]

    Replacing with the information that we know for part two that we start from [itex]t_1[/itex], position [itex]x_1[/itex] and initial velocity [itex]v_1[/itex] so replacing with these information:

    [itex]x (t) = x_1 + v_1 (t-t_f) - \frac{1}{2} r (t-t_f)^2[/itex]

    [itex]v (t) = v_1 - r (t-t_f)[/itex]

    Also we know that at final time [itex]t_2 = t[/itex] the movil finish with velocity [itex]v_2 = 0[/itex] at the point [itex]x_2 = s[/itex]. You have a system of ecuations im sure you are able to solve, give it a try.
     
  4. Jul 11, 2012 #3
    Very interesting but i don't know how to prove it.
    It looks like the equivalent of the 2 acceleration value analogous to effective K of 2 springs connected in series.

    Keffective=k1k2/(k1+k2)

    In the above example, the effective acceleration is equal f1f2/(f1+f2)
    Maybe same approach can be taken to prove as calculation to find effective K of springs connected in series.
     
  5. Jul 12, 2012 #4

    TSny

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    Let t1 be the time of the acceleration phase and t2 the time of deceleration. The increase in speed during the acceleration phase must equal the decrease in speed during the deceleration phase. Hence, ft1=rt2.

    Therefore, t2 = (f/r)t1.

    Thus, show t = t1 + t2 = [itex]\frac{f+r}{r}[/itex]t1. Therefore, t1 = [itex]\frac{r}{f+r}[/itex]t and t2 = [itex]\frac{f}{f+r}[/itex]t. (These are expressions for t1 and t2 in terms of the total time t.)

    Apply the general formula x = vot + [itex]\frac{1}{2}[/itex]at2 to the acceleration phase and then to the deceleration and note that the initial velocity of the deceleration phase is the final velocity of the acceleration phase = ft1.

    s1 = [itex]\frac{1}{2}[/itex]ft12

    s2 = (ft1)t2 - [itex]\frac{1}{2}[/itex]rt22

    Try substituting previously derived expressions for t1 and t2 in terms of the total time t and see if you can simplify s = s1+s2 to the desired result.

    [Maybe there's a better way.]
     
    Last edited: Jul 12, 2012
  6. Jul 12, 2012 #5
    I think its along these lines. The final velocity in the first part of the journey is the same as the initial velocity in the second part of the journey. These yields the equality f*s1 = r*s2 for v^2 given that s1 + s2 = s . So part of the approach might be using the kinematic equation that does not use time to begin with. And then substituting with time and then maybe too much algebra for my taste.
     
  7. Jul 12, 2012 #6

    TSny

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    Good suggestion!
     
  8. Jul 12, 2012 #7
    TSny

    Looked at it at lunch. It can be done with your ratio without going "backwards". But its a royal pain.

    I guess this was a practice your substitutions and algebra problem. Maybe its easier than what I am looking at, but if not, it was a cruel form of torture and quiet possibly violates the US constitution.
     
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