# Scientific Commuting: How Much Faster?

Back in the summer, I did a post mathematically comparing two routes to campus, one with a small number of traffic lights, the other with a larger number of stop signs, and looked at which would be faster. Later on, I did the experiment, too.) Having spent a bunch of time on this, I was thinking about whether I could use this as a problem for the intro physics class. I decided against it last fall, but something else reminded me of this, and I started poking at it again.

So, I played around a bit with some numbers, and came up with the following possible framing for a question. I’ll throw this out here, to see what people think, then I’ll post some solutions in a few hours.

A car starts from rest at the beginning of a straight 1km course, accelerates up to some speed, cruises at constant speed for a while, then decelerates to a stop at the end of the course. A second, identical car does the same course, but decelerates to a stop at the halfway point. It then immediately accelerates back to its cruising speed, and then decelerates to a stop at the end of the course.

How much faster does the second car have to go in order to complete the course in the same time as the first car?

Obviously, this requires some assumptions about the speed and acceleration of the car, which might make for some useful discussion. You can find the necessary math at the first link above, if you want to work it out. For a class using a clicker-type response system, I’d turn it into a poll question like so:

So, we’ll leave this up for a while, and then this afternoon, I’ll post some analysis. Feel free to share your thoughts about appropriate assumptions, etc. in the comments. Or to suggest ways to re-cast the problem that might be more useful. This is classical physics, though, so you don’t get to choose more than one option.

## 5 thoughts on “Scientific Commuting: How Much Faster?”

1. Clay B says:

For very slow top speeds, the time lost in the middle is very small, so the lower bound is 1.

I get 1.5 as an estimate of the upper bound by comparing the distance covered (area under the curve in a plot of velocity vs. time) for the two cars when the second car just reaches top speed for an instant (triangle wave vs. trapezoid wave) but that’s only exact if the first car is allowed to accelerate 1.5 times as fast as the second car.

2. Susan B. says:

I solved it by drawing some pictures of velocity versus time (assuming both cars have the same (constant) acceleration and deceleration, though I’m not sure it matters if the acceleration and deceleration have different magnitudes). I drew a trapezoid for the first car, and two trapezoids for the second, then tried to guess how much higher the second car’s trapezoids need to be to get the same area (distance traveled).

If the first car spends a large proportion of the total time traveling at its top speed, then the second car’s top speed is very close to the first: (1+epsilon) times the first.

At the other extreme, if the second car travels such that it accelerates to its top speed and then immediately decelerates, then does the same again (so that it only travels at its top speed for an instant), then we get our upper bound: looking at my drawing, I think it’s sqrt(2) times the first car’s top speed, but I haven’t actually worked the numbers and I may be wrong.

Clearly there are some scenarios for which there is no answer. In order for the second car to be able to keep up at all, it is necessary for the first car to spend more than half the total time traveling at its max velocity (how much more, I haven’t yet worked out).

To answer the larger question, I think this is an excellent problem and I’ll probably be stealing it to see if I can work it into my calculus classes!

3. Tom Singer says:

For the first car, I’ll assume it cruises at 60 mph, which is conveniently right about 100 km/h. And I’ll assume it accelerates uniformly from 0-100 km/h in 8 s. Reasonably sporty, but not something you want to bring to a drag race. (The linear assumption is not great, but it’s probably not horrible up to 60 mph, either.)

Average velocity in those 8 s is 50 km/h, so you drive a distance of d1 = 50 km/h * 8 s * 1 h / 3600 s = 400 km / 3600 = 1/9 km, or about 0.11 km in 8 s.

Assume deceleration performance is similar – 100 to 0 in 8 s. (Neglecting drag, it seems like braking should be quite linear.) So, the acceleration and deceleration are going to take up 16 s and 0.22 km. The remaining 0.78 km is traveled at 100 km/h, so it takes 0.78 km / 100 km/h = 0.0078 h * 3600 s/h ~= 3/4 * 36 = 27 s. So, total time is T = 43 s.

For car 2, assume it has the same uniform acceleration and deceleration, a = 100 km/h / 8 s = 12.5 km/h/s. (Depending on what the necessary velocity is, linear acceleration might be a much worse assumption here.)

Is it even possible to cover 1 km in 43 s, if you accelerate and decelerate the whole way there? Equivalently, how fast can you cover 1/4 km if you accelerate at 12.5 km/h/s? It needs to be faster than 43/4 s = 10.75 s.

d = 1/2 a * t^2
t = â(2d / a)
t = â(1/2 km / 12.5 km/h/s)
t = â(1/25 hs)
t = 1/5 â(hs * 3600 s/h)
t = 1/5 â(3600 s^2)
t = 1/5 * 60 s
t = 12 s

So… not possible for car 2 to cover the km in the same time as car 1.

4. Tom Singer says:

That integral sign should be a square root. That’s odd.

Anyway, it’s reasonable to ask what the max speed car 1 can cruise at that will allow car 2 to complete the course, assuming a = 12.5 km/h/s. That means car 1 needs to take T = 4 * 12 s = 48 s to complete D = 1 km.

The acceleration and deceleration will each take t1, leaving T – 2t1 = t2 for cruising. The distance traveled while accelerating and decelerating is d1 = a * t1^2 / 2, leaving D – 2d1 = d2 for cruising. So, the cruise velocity is v = d2 / t2. And that velocity is also v = a t1.

d2 / t2 = a t1

D – 2 * a t1^2 / 2 = d2

(D – a t1^2) / (T – 2t1) = a t1
D – a t1^2 = a t1 (T – 2t1)
D – a t1^2 = a t1 T – 2a t1^2
a t1^2 – a T t1 + D = 0
t1 = [aT Â± sqrt(a^2 T^2 – 4aD)] / (2a)

v = [aT Â± sqrt(a^2 T^2 – 4aD)] / 2

aT = 12.5 km/h/s * 48 s = 600 km/h

(aT)^2 = 360000 km^2/h^2

4aD = 4 * 12.5 km/h/s * 1 km
4aD = 50 km^2/h/s * 3600 s/h

sqrt() = sqrt(360000 – 180000) km/h
sqrt() = sqrt(180000) km/h
sqrt() = 100 sqrt(18) km/h
sqrt() = 300 sqrt(2) km/h

v = [600 km/h Â± 300 sqrt(2) km/h] / 2
v = 300 km/h Â± 150 sqrt(2) km/h
v = 512 km/h or v = 88 km/h

But if v = 512 km/h, t1 = a / v is greater than T = 48 s; we need it to be at most T/2. So that’s not a valid solution, and we’re left with v = 88 km/h.

Since car 2 is reaching a max velocity after 12 s of acceleration at 12.5 km/h/s, that’s 150 km/h, or 1.5 times the velocity of car 1. I suspect that if I were more rigorous with my symbolic representation, acceleration would drop out of this upper limit entirely, which would leave me with Clay B’s solution.

5. Joffan says:

What velocity does driver 2 regard as a “stop”?

😉

[Nice work Tom.]