So, the previous post poses a physics question based on some previous fooling around with modeling my commute:

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?

This is deliberately somewhat vague, because I want to leave room for some discussion of what are reasonable values for cruising speed and acceleration, and so on. For the purposes of attempting to answer this, I used values that were moderately realistic and also mathematically convenient: A cruising speed for the first car of 16 m/s, and an acceleration (both speeding up and slowing down) of 4 m/s/s.

So, how do you work this out mathematically? Well, if you look back at the first post I did modeling this, you find that the time required to make the trip in N stages is:

i-e4fa5c97bb90d65f1d3ee699bf310ff0-commute5.png

Here you see the reason why I didn’t end up doing anything with this last fall: this is an equation that’s a little tricky for intro physics students, whose level of math preparation isn’t always what we would like it to be. To get an algebraic solution, you would need to take this equation with N=1 and set it equal to the equation for N=2 but a different final speed, and solve for that final speed. That’s more math than I really want to wrestle with, and I guarantee it would be a disaster for most of our intro students.

However, this is an easy enough equation to plug into a computer and generate solutions, which you can plot as nice graphs. So here are a few graphs giving the answer for the situations I looked at:

i-60195e891113e31313f1135d2cd07879-16msDrivingSpeed.PNG

The horizontal line here indicates the time to cover 1km for a car accelerating at 4m/s to a speed of 16 m/s. The points represent the time to complete the trip (vertical axis) for a car with a given cruising speed (horizontal axis). When the two cross each other, that’s the speed at which the travel times are equal, which turns out to be between 17 and 18 m/s.

That was a little surprising to me, as it’s only about a 10% increase in the speed. The difference gets bigger if you increase the cruising speed of the first car to 25 m/s:

i-a36867e8bcf1f2e04cffed145cebe23b-25msDrivingSpeed.PNG

In this case, the speed for the second car is about 35 m/s, or 40% faster. You can also see from this that there’s a minimum in the time as a function of speed (which you can find with calculus, if you care), and if you increase the initial speed enough, that minimum will move above the time for the first car, at which point there’s no speed you can accelerate to that will let you catch the first car.

So, the correct answer to the poll depends on what your values for speed and acceleration are. I punched a bunch of different numbers in, though, and didn’t seem to find any where the difference in speeds was greater than about 40% (though, admittedly, I didn’t spend that much time on this….).

The question doesn’t ask this, but you can play with other factors as well. If you increase the number of stops, you increase the required speed for the second car:

i-29a10517a7a4c07c9308cf9f17e00469-16msDriving4stop.PNG

This graph is for doing the trip in four segments, that is, with three stops along the way. Again, there’s a minimum time for a particular speed, and this minimum value moves up as the number of stops increases. For five or more segments (four stops), there’s no speed at which the second car catches the first.

You can also vary the length of the course:

i-3bdbd5f79b13acb59875e5cf519a3cfb-16msDrivingSpeed500m.PNG

This is for a 500m course rather than 1km. This also increases the difference in speeds– the second car needs to be going at 20-21 m/s to match the first car at 16 m/s. (This makes sense– you’ve got less room in which to make up the ground you lose while stopping…)

I suspect this still probably requires more mathematical sophistication than we can rely on in our intro courses, but it’s kind of fun to play around with. And it was surprising to me to see how small the difference was for reasonable values, so whether it ends up being useful in class or not, I learned something. And got two blog posts out of the deal, so, hey, big win for me…

Comments

  1. #1 excitedstate
    March 23, 2012

    My intuition was right, for the speeds I had in mind. I don’t think you’d find stop signs on roads where the speed limit is much above 40 mph, in which case the presence of a stop sign doesn’t actually slow you down all that much.

    Of course, I tend to accelerate more gradually when the speed limit is lower, so that would affect the result in practice.

    Also, you have a unit error below the first graph, where you give acceleration in m/s.

  2. #2 Rosie Redfield
    March 23, 2012

    No need for any fancy math, just consider the extremes:

    1. Acceleration/deceleration is infinitely fast. Ratio of speeds = 1.

    2. Acceleration/deceleration is very slow, just fast enough that car 2 reaches cruising speed, then immediately begins decelerating. Car 1′s average speed is 1.5 times that of car 2.

  3. #3 joemac53
    March 24, 2012

    We would do similar problems with a velocity/time graph, using the area under the curve as the distance traveled. Since distances are the same, areas are equal. Once we made some assumptions about acceleration we could do most straightforward problems. One assumption that we always argued about was how much friction you could get between tires and road. We finally decided that 1 g was the upper limit, although I am still not certain that is the case.

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