Suppose that you, law-abiding driver that you are, are going along the highway at the posted speed limit of 70 mph. You see in your rear-view mirror a car traveling at high speed in the lane next to yours. Suppose that car is traveling at 100 mph. At the instant that the other car is right next to you, you both see an obstruction ahead, say a tree lying across the road. You both slam on your brakes to the maximum and you come to rest just before you hit the barrier. What would be the speed with which the other car hits the tree?

Making some simple assumptions, such as that both cars and drivers are very similar, this problem becomes an elementary physics problem that anyone who knows basic kinematics can solve. Even if you don’t you can make a guess. The point of the problem is that the result was quite surprising, even for me. It would be a good problem to give introductory physics students, both as an application of physics and to alert them to the dangers of speeding.

This video gives the solution and different ways of arriving at it.

(Via Mark Frauenfelder.)

Ogvorbis wants to know: WTF!?!?!?! says

Were I in my car (14 Elantra) and Boy were in his car (15 Mustang), he would stop 30 or more feet sooner than I. His car weighs a little more, but his brakes are freakin’ huge and his tyres are twice the width of mine, plus they are performance tyres (Pirelli Scorpions).

Still going to work this through with him to get him to slow down. Of course, his car is orange, not red, but he’ll still get the point.

And I ballparked it and came up with less than 30 miles per hour. I completely spaced the idea of using kinetic energy to come up with the proportions.

At my park, we are working on a math olympics for high school students. And this one could definitely be used to show how fast trains stop. And the fact that the trains in question weigh in at, say, 3,000 tons, wouldn’t matter.

DonDueed says

At first I made the assumption that both vehicles would decelerate at the same rate (same dv/dt), and since the delta-t was the same, so would the delta-v be (thus yielding a 30mph collision).

But that doesn’t work, because the red car is going

faster. That means it gets to the barrier in less time. That’s why the delta-v is so much less.Rob Grigjanis says

DonDueed@2: Right. Assuming constant deceleration a, origin x=0, and starting withx(t) = (v0)t -- (1/2)at^2

You can show that

x(t) = (1/2a)((v0)^2 -- v^2)

Same delta-x for both cars, so the difference in v^2 is the same for both cars.

Rob Grigjanis says

‘v’ should be ‘v(t)’.

Rob Grigjanis says

Forgot to include the velocity equation

v(t) = v0 -- at

which can be used to substitute (v0 -- v)/a for t in the distance equation.

Rob Grigjanis says

It just occurred to me that a simpler solution is noting that the work done in braking is

Force x distance = mad

This would be the same for both cars (same m, same a, same d), and it would equal the change in kinetic energy for each. So

Δ(v1^2) = Δ(v2^2)

And I see that the bloke in the video sort of uses this, with some handwaving.

flex says

Somewhat related, I’ve on occasion discussed the differences in winter driving in Michigan snow with recent immigrants who’ve never had the experience.

Since these are engineers, I remind them of the equation for momentum. “I understand momentum”, they invariably tell me. “Okay” I reply, “Then apply it to this situation. Your car is the same weight as mine, but I’m going 30 MPH and you are going 60 MPH. Assume that when we both hit black ice that friction is negligible. What is the difference in momentum.”

All of a sudden a little light comes on in their eyes, and they realize what increasing by a square really means.

hyphenman says

Mano,

Great video (and problem). This was the kind of challenge that kept me in high school physics class. (Our final was to derive the period of a double pendulum.)

One aspect of physics equations that has always fascinated me is how exponents are nearly always whole numbers.

That Einstein’s most famous equation would be E=MC^2 and not E=MC^1.9 or E=MC^2.1 has always seemed to me to be seductive.

Cheers,

Jeff

Matt G says

Hyphenman@8-

Squares are not uncommon in math and science. Think about derivatives in calculus and the inverse square law found in many branches of physics (like gravity -- when you halve the distance between masses, you quadruple the force).

hyphenman says

@Matt G, No. 9,

Yes, but the fact that squares are not uncommon (outside the figuring of the area of a space) is, for me, a fascinating reality.

I’m not questioning that reality in any way, just saying that given the infinity involved, when whole numbers appear I have to marvel at how the universe works.

Cheers…

Rob Grigjanis says

hyphenman@10: The fact that integer powers are not uncommon is not really that marvellous; it is intimately related to the number of spatial (or spacetime) dimensions. A non-integer number of dimensions would be really weird.