Thread by @Dan_Schwartz

@DataRepublican, an actual Georgia Tech Engineer here:
Even without resorting to quantum physics, solely using Newtonian physics and taking into account significant digits the way you presented the quiz, there is actually a range of correct answers!

Let me explain 1/🧵

OK, @DataRepublican presented this problem:
A traveler must make a 60-mile round trip between two towns, Aliceville and Bobtown.

The distance each way is 30 miles. Going from Aliceville to Bobtown, the traveler drives at exactly 30 miles per hour.

By the time they reach Bobtown, they decide they want to average 60 miles per hour for the entire 60-mile journey. 2/🧵

However, let's rewrite the problem taking into account the measurement tolerances due to significant digits:

A traveler must make a (6 x 10¹) mile round trip between two towns, Aliceville and Bobtown.

The distance each way is (3 x 10¹) miles. Going from Aliceville to Bobtown, the traveler drives at exactly (3.0 x 10¹) miles per hour.

By the time they reach Bobtown, they decide they want to average (6 x 10¹) MPH for the entire (6 x 10¹) mile journey.
3/🧵

Now, how many miles are really in the round trip between Aliceville and Bobtown?

6 x 10¹ miles when resolved to significant digits as @DataRepublican presented it can be anywhere between:
55 miles < 6 x 10¹ miles < 65 miles.
4/🧵

Repeating for velocity ("speed" if you never learned vector-based Newtonian physics):

"Going from Aliceville to Bobtown, the traveler drives at exactly (3.0 x 10¹) miles per hour"

This gives us a velocity range for Aliceville→Bobville of:
29 mph < ν < 31 mph
5/🧵

Recapping where we are,
• 55 miles < Aliceville→Bobville→Aliceville < 65 miles;
• 29 mph < Aliceville→Bobville < 31 mph.

This is simply a measurement tolerance issue with two independent variables, that we were taught in freshman chemistry to illustrate the principle. 6/🧵

@DataRepublican Just to get our arms around the problem, since
55 miles < Aliceville→Bobville→Aliceville < 65 miles,
let's take the 55 mile figure.

This means with ν = 31 mph for Aliceville→Bobville, the time actually used is:

(55÷2) miles
——————— ≈ 0.89 hours ≈ 53:40
31 miles/hour

7/🧵

Having used 53:40 to cover (55÷2 = 27.5) miles for Aliceville→Bobville, that leaves us at most 6:40 (400 seconds) to cover the 27.5 miles for the Bobville→Aliceville return trip.

27.5 miles ÷ 400 seconds = .0688 miles/sec ≈ 250 mph or faster, depending on time, distance and velocity tolerances..

OK @DataRepublican, prove me wrong.😉
8/end 🧵