Constant Acceleration - A Special Case — Question 9

Question 9

Equal distances need not take equal times

A cart moves east with one constant positive acceleration. Two successive track segments are each 10 m long. In a first proposed model it starts from rest at the first segment’s entrance and takes exactly 2 s to traverse that segment.

1. Determine acceleration in the proposed model. Find the total time to cover both segments and the time spent on the second one.

2. Explain why assigning 2 s to each equal-distance segment contradicts the model. Compare the velocities at the first and second segment exits.

3. Now replace the release-from-rest claim with a new exact record: first 10 m in 2 s, next 10 m in 1 s. Retain one constant acceleration but allow an unknown initial eastward velocity $u$. Solve for $u$ and $a$ using positions at 2 and 3 s.

4. Do the new times alone force acceleration to vary? State precisely which assumption from the first model they contradict.

Question 9 - Solution

Equal distances need not take equal times

(a) Use elapsed time from the same release event.

The first segment gives $10=\tfrac12a(2)^2$, so $\boxed{a=5\ \mathrm{m/s^2}}$. To reach 20 m from rest, $$20=\tfrac12(5)T^2,\qquad\boxed{T=2\sqrt2\ \mathrm{s}}.$$ The second segment therefore takes $\boxed{2(\sqrt2-1)\ \mathrm{s}\approx0.8284\ \mathrm{s}}$.

(b) The second segment begins with an already positive velocity.

The exit velocities are $\boxed{v_1=10\ \mathrm{m/s}}$ and $\boxed{v_2=10\sqrt2\ \mathrm{m/s}}$. The cart covers the second equal distance at larger velocities, so it needs less time. Resetting both time and velocity to zero at the middle gate would describe a restart that is absent from the model.

(c) Let the data determine the initial velocity rather than assuming rest.

$$\begin{aligned}2u+2a&=10,\\3u+\tfrac92a&=20\end{aligned}\quad\Rightarrow\quad\tfrac32a=5,\quad a=\tfrac{10}{3},\quad u=\tfrac53\quad\text{(SI)}.$$

For the replacement record, $$10=2u+2a,\qquad20=3u+\tfrac92a.$$ The first equation gives $u=5-a$. Substitution into the second yields $20=15+\tfrac32a$, hence $$\boxed{a=10/3\ \mathrm{m/s^2}},\qquad\boxed{u=5/3\ \mathrm{m/s}}.$$ The velocities at 2 and 3 s are $25/3$ and $35/3\ \mathrm{m/s}$. Their mean on the 1 s second segment is 10 m/s, giving exactly the required 10 m. The velocity remains positive throughout.

(d) Reject only the assumptions that the evidence contradicts.

The replacement timings are incompatible with the combined constant-acceleration and start-from-rest hypothesis: that hypothesis predicted about 0.8284 s for the second segment. They do not force variable acceleration, since the explicit nonzero-initial-velocity constant-acceleration solution fits them exactly. A failed prediction can identify a problem in a combination of assumptions without uniquely identifying which assumption was wrong.