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An Example

A train on a single line track is told to be at position $x_{res}$, at time $t_{res}$, and to be moving at exactly $v_{res}$ when it hits that point. Where $_{res}$ denotes the reservation.

The train driver knows the current time ($t_{cur}$), it's current speed ($v_{cur}$) and it's current position ($x_{cur}$). Where $_{cur}$ denotes the current value.

Provide an equation that gives the acceleration the train driver must apply at $t_{cur}$, in order to be on track to meet the reservation.

The train could need to accelerate to faster than $v_{res}$ then slow down. It could instead have to constantly accelerate, or even maintain the exact same speed. It all depends on the situation.

I'm looking to determine acceleration as a function of time, in order to meet a space-time-velocity reservation. - That is: be at location $x$, at time $t$, with final velocity $v$.

The problem domain can be considered 1 dimensional.

Available Values

  • Current and final location ($x_1$ & $x_2$), thus $\Delta x$.
  • Current and final time ($t_1$ & $t_2$), thus $\Delta t$
  • Current and final velocity ($v_1$ & $v_2$), thus $\Delta v$

I am aware there are likely infinite acceleration curves that would technically solve this problem, an ideal solution would result in a curve with the least extreme accelerations.

The final answer should be an equation giving acceleration using the available values above. I would appreciate it if you could explain how you found your answer, and your patience with the limited knowledge of a laymen.

What I've tried

I'm familiar with equations of motion under constant acceleration, however varied acceleration like this is still a little over my head. I have posted this question on Reddit, and received an answer, however that too is over my head, and I am unable to work through it, though I have attempted to. You can see this post and my attempts at working through it here

I am not a Physicist or a Mathematician, so please excuse any mistakes or misconceptions on my part. I would be truly grateful for any help.

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1 Answer 1

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You can't satisfy all four conditions

$$x(t_1)=x_1\tag 1$$ $$\dot{x}(t_1)=v_1\tag 2$$ $$x(t_2)=x_2\tag 3$$ $$\dot{x}(t_2)=v_2\tag 4$$

with a constant acceleration. But you can with a linearly changing acceleration of the form

$$\ddot{x}(t)=A+B(t-t_1).\tag 5$$

Integrating (5) and using condition (2) to determine the constant of integration, the velocity is

$$\dot{x}(t)=v_1+A(t-t_1)+\frac12B(t-t_1)^2.\tag 6.$$

Integrating (6) and using condition (1) to determine the constant of integration, the position is

$$x(t)=x_1+v_1(t-t_1)+\frac12A(t-t_1)^2+\frac16B(t-t_1)^3.\tag 7$$

Imposing condition (3) on (7) and condition (4) on (6) leads to two simultaneous equations to solve for $A$ and $B$. The solution is

$$A=2\frac{3(x_2-x_1)-(v_2+2v_1)(t_2-t_1)}{(t_2-t_1)^2}\tag 8$$

and

$$B=-6\frac{2(x_2-x_1)-(v_2+v_1)(t_2-t_1)}{(t_2-t_1)^3}.\tag 9$$

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  • $\begingroup$ Where do the equations for A and B come from? Are they known equations with a source I could reference in a paper? $\endgroup$
    – Torantula
    Commented Feb 25, 2020 at 14:57
  • $\begingroup$ And come to think of it, same goes for the final equations. $\endgroup$
    – Torantula
    Commented Feb 25, 2020 at 14:57
  • $\begingroup$ Also, I see you've used $t$, $t_1$ and $t_2$, which seems to suggest there is some initial time value, however the only values available are the current and final times. The same goes for velocity and position. $\endgroup$
    – Torantula
    Commented Feb 25, 2020 at 15:27
  • $\begingroup$ There has to be something happening between what you call the “current” time and the final time. Consider $t_1$ to be the “current” time, $t_2$ to be the final time, and $t$ to be an arbitrary time in between these. $\endgroup$
    – G. Smith
    Commented Feb 25, 2020 at 16:11
  • $\begingroup$ I have edited to answer to explain the steps in detail. However, if you do not understand calculus, it will make no sense because you have to use integration to get the velocity from the acceleration and the position from the velocity. $\endgroup$
    – G. Smith
    Commented Feb 25, 2020 at 16:27

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