Am I making the right assumption about a jump discontinuity in the acceleration?

Question

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?

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.