A train heads from Station A to Station B, 4 km away. If the train begins at rest and ends at rest, and its maximum acceleration is 1.5 m/s^2 and maximum deceleration is -6 m/s^2, what's the least time required to complete the journey?

My attempt at a solution: I assumed the acceleration function, $a(t)$, has a jump discontinuity at some time, say $t_1$, where it jumps from 1.5 to -6. Under this assumption, it's easy to use the velocity boundary conditions to get the total time $t_f$ is $\frac{5}{4}t_1$, because the velocity $v(t)$ for $t>t_1$ is just $1.5t_1-6(t-t_1)=7.5t_1-6t$. Setting this equal to 0 at $t_f$ gives this result.

The position $x(t)$ for $t\leq t_1$ is just $\frac{1}{2}(1.5)t^2$ and for $t>t_1$ it's just $\frac{1}{2}(1.5)t_1^2+(1.5t_1)(t-t_1)-\frac{1}{2}(6)(t-t_1)^2$. Using the fact that $x(t_f)=4000$, and substituting $t_f=\frac{5}{4}t_1$, I find $t_f\approx 86.06$ seconds.

But my teacher's solution says the minimum time is $81.65$ seconds. Who's wrong? Is my assumption about the jump discontinuity of $a(t)$ what's wrong?

  • $\begingroup$ Hi symplectomorphic, and welcome to Physics Stack Exchange! I changed the formatting of your question a bit to make it easier to read. $\endgroup$
    – David Z
    Sep 26, 2012 at 3:38
  • $\begingroup$ I think your change of title is misleading -- it's a much less interesting question. The question is a brachiostone-style time minimization problem: the point of putting "time minimization" in the title was to bring out this connection. $\endgroup$ Sep 26, 2012 at 3:39
  • $\begingroup$ brachistochrone* $\endgroup$ Sep 26, 2012 at 3:41
  • $\begingroup$ Ah, well, that didn't seem clear to me. Feel free to change the title (or more of the post) to reflect what you wanted it to ask. (It's highly recommended, but not required, to make the title a question that corresponds to the main thing you are asking.) $\endgroup$
    – David Z
    Sep 26, 2012 at 3:52

1 Answer 1


I agree with you that the distance travelled up to the changeover point is:

$$ x = \frac{1}{2}a_1\left(\frac{4t_f}{5}\right)^2 $$

but what I would do next is start from the other end and point out that:

$$ 4000 - x = \frac{1}{2}a_2\left(\frac{t_f}{5}\right)^2 $$

adding these together and rearranging gives:

$$ t_f = \sqrt{\frac{50 \times 4000}{16a_1 + a_2}} \approx 81.65 $$

so your teacher is indeed correct (it had to happen one day :-). Your equation for the total distance:

$$ 4000 = \frac{1}{2}(1.5)t_1^2+(1.5t_1)(t-t_1)-\frac{1}{2}(6)(t-t_1)^2 $$

looks fine to me, so I'd guess you just made a mistake in the algebra.

  • 1
    $\begingroup$ Found my algebra mistake after reading through my work for about the fiftieth time. Thanks for confirming I wasn't losing my mind about the general approach. $\endgroup$ Sep 26, 2012 at 6:42

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