Thanks once again to Car Talk for giving me such great questions.
In the last episode, a caller asked about stopping a car. He said that when he stops, he turns the car left and right to increase the total distance over which he stops but decrease the linear stopping distance. Tom and Ray point out that this practice is a very bad idea and they don't think it would even theoretically work. So, will it?
Ok, model time. I think I already know the answer to this question, but the model might be the most convincing answer. How will I model it? With python, of course. But just to make things fun, let me use the following situation for the turning-topping car:
- Car with a mass of 1200 kg.
- Coefficient of static friction between the tires and the road of 0.8
- Initial car speed of 70 mph (31 m/s).
- I will assume the car turns in a circle of radius 15 meters until it is 10 degrees off the original line then turns back.
There is another assumption. I will assume that the magnitude of the frictional force is constant. So, as the car is turning, a component of the frictional force will be used to turn the car and the rest will be there to slow it down. Here, this diagram will help. This shows the car turning and stopping at three different times.
Here, the blue arrow represents the total frictional force. I have broken this frictional force into two components. The green arrow represents the component of friction needed to make the car turn. The red arrow represents the component of friction in the opposite direction as the velocity vector. This red-labeled component of friction slows the car down.
Maybe you can already see the problem. When you turn, you have to use part of your frictional force for turning instead of slowing down. So, although you might have more distance to stop, you will have less force stopping the car.
Ok, now for the model. Here is my numerical "recipe":
- Calculate the total frictional force (really, you only have to do this once).
- From the velocity, calculate how much of this frictional force has to point perpendicular to the velocity of the car (you know - centripetal acceleration). Note that I will adjust this amount of turning so that the radius of the turn (at that time) will be very close to the smallest radius possible. You can't just turn in a circle of whatever radius you want because the frictional force has some maximum value.
- Find out the left over component of friction that will be in opposite direction as the velocity.
- Now that I have the force as a vector - apply the momentum principle.
- Use the momentum to update the position.
- Repeat.
So, does it work? For this first case, I am going to have a car that is turning and braking. It will turn at a constant radius - starting with a radius that is 1.25 the minimum radius for the starting speed. Here is a plot of the path for that car along with a car stopping in a straight line.
For this case, the straight driving car stops in 61 meters but the turning car stops in an x-distance of 68.5 meters. The total distance traveled for the turning car was 71 meters.
Ok, there are lots of different ways to stop a car. I could turn just to the left with a constant radius (like above) or it could take a sharper turn as the car slows down. Another option would be turn back and forth and not just to one side. Let me go ahead and run all these cases.
Notice that in all of these cases, the swerving or turning car stops in a greater x-distance. Of course the total distance is also greater, but that was the original idea. Let me make a plot just showing the two swerving.
Here the blue track is a car that turns with a radius of 3 times the minimum turning radius. It stops in an x-distance of 64.8 meters (the straight stopping car stopped in 61 meters). The red line represents a car that turns with a smaller radius - 1.25 times the minimum turning radius. It stops in an x-distance of 95 meters. So both of these turning cases do indeed increase the total distance over which the car stops. They do stop over a longer distance. They do not stop in a shorter horizontal distance.
Momentum
If you like, you could think of this problem in terms of momentum. Let me call the original direction of motion the x-direction. In this case, whatever the car does it needs to take its component of momentum in the x-direction down to zero. The momentum principle in just one dimension says:
If the car swerves instead of stopping in a straight line, clearly the x-component of the force (friction) will be smaller at times since you have to use part of this frictional force to turn. The problem is that this expression also has time in it. I remove the time by looking at the definition of average velocity:
So I can write the change in x-momentum as:
Now back to a swerving and straight stopping car. If the swerving car has a constant x-acceleration (which could be approximately true if the swerves were quick) then both cars would have the same starting and stopping x-velocity and the same average velocity. The cars would also have the same change in x-momentum. The swerving car, however, would have a smaller x-component of force so Δx for that swerving car would also have to be larger.
The answer
If you are here just for the answer, it seems clear. The best way to stop is by staying in a straight line. Do not swerve, do not pass go. Actually, I just realized I looked at a very similar problem before - should you turn or stop to avoid hitting a brick wall.
