Constant Acceleration - A Special Case — Question 7

Question 7

Bound acceleration from uncertain gate speeds

A point moves east from gate A to gate B with one constant acceleration and no reversal between gates. True initial speed $u$, final speed $w$, and gate separation $L$ obey independent hard bounds: $$u\in[7.9,8.1]\ \mathrm{m/s},\quad w\in[3.9,4.1]\ \mathrm{m/s},\quad L\in[11.8,12.2]\ \mathrm{m}.$$ Every combination within these bounds is allowed before applying the model. Use $$w^2=u^2+2aL,\qquad L=\tfrac12(u+w)T.$$

1. Use the central values to estimate acceleration and transit time. Explain why the acceleration is negative although both velocities are eastward.

2. Find the tight interval for $a$. State which choices of $u,w,L$ give its two endpoints; do not treat negative numerators as positive when choosing denominator extremes.

3. Find the tight interval for transit time. Do the combinations that extremize time have to be the combinations that extremize acceleration?

4. Is every allowed input triplet compatible with positive transit time and no reversal in this constant-acceleration model? Explain why these bounds are conditional on the model rather than a description of arbitrary hidden motion.

Question 7 - Solution

Bound acceleration from uncertain gate speeds

(a) Use signed velocity changes on the declared axis.

At $u=8,w=4\ \mathrm{m/s}$ and $L=12\ \mathrm{m}$, $$\boxed{a=(16-64)/24=-2\ \mathrm{m/s^2}},\qquad \boxed{T=24/(8+4)=2\ \mathrm{s}}.$$ The velocity remains eastward but decreases in magnitude, requiring westward acceleration.

(b) Bound the negative ratio carefully.

All allowed numerators $w^2-u^2$ are negative. The most negative ratio uses largest $u$, smallest $w$, and smallest $L$; the least negative uses smallest $u$, largest $w$, and largest $L$. Thus $$\boxed{a\in\left[\frac{3.9^2-8.1^2}{2(11.8)}, \frac{4.1^2-7.9^2}{2(12.2)}\right] =\left[-\frac{126}{59},-\frac{114}{61}\right]\ \mathrm{m/s^2}}.$$ The exact endpoints are approximately $-2.1356$ and $-1.8689\ \mathrm{m/s^2}$. These feasible corner choices make the interval tight.

(c) Extremize the time ratio independently.

Since $T=2L/(u+w)$ has a positive numerator and denominator, $$\boxed{T\in\left[\frac{2(11.8)}{8.1+4.1},\frac{2(12.2)}{7.9+3.9}\right] =\left[\frac{118}{61},\frac{122}{59}\right]\ \mathrm{s}}.$$ This is approximately $[1.9344,2.0678]\ \mathrm{s}$. The speed choices differ from the acceleration-extreme choices. Bounds on different derived quantities need not be attained by one common input triplet.

(d) Check feasibility rather than only arithmetic extremes.

$$v(t)=\left(1-\frac tT\right)u+\frac tT w>0\quad(0\leq t\leq T),\qquad a=\frac{w-u}{T}.$$

For every triplet, $T>0$ and $a=(w-u)/T<0$. The resulting velocity varies linearly between two positive endpoints and cannot reverse. Its displacement equals $(u+w)T/2=L$, so every input triplet has a compatible trajectory. Arbitrary time-varying accelerations could give different transit times or local accelerations; the results rely on the stipulated single constant-acceleration model.