This is a classic problem. You are in a car heading straight towards a wall. Should you try to stop or should you try to turn to avoid the wall? Bonus question: what if the wall is not really wide so you don't have to turn 90 degrees?
Assumption: Let me assume that I can use the normal model of friction - that the maximum static friction force is proportional to the normal force. Also, I will assume that the frictional coefficient for stopping is the same as for turning.
I am going to start with the case of trying to stop. Suppose the car is moving towards the wall at a speed v0 and an initial distance s away from the wall. Diagram time:
This is a 1-d problem. So, let me consider the forces in the direction of motion. There is only one force - friction. Now - you might be tempted to use one of the kinematic equations. Well, I guess that is just fine. The following equation is appropriate here.
Really though, I would think - hey distance. That means use the work-energy equation. It gives you the same thing though - essentially. Since I already started with this kinematic equation, let me proceed. In the direction of motion, I get:
Putting this into the above kinematic equation (with the change in x-distance as just s). Oh, note that I am using the maximum static frictional force. I am assuming that this will be the shortest distance you could stop. Also, I am assuming that I the car stops without skidding.
There you have it. That is how far the car would need to stop. Quick check - does it have the right units? Yes.
Now, how far away could the car be and turn to miss the wall? Really, the question should be: if moving at a speed vo, what is the smallest radius turn the car could make?
For an object moving in a circle, the following is true:
Here is my review of acceleration of an object moving in a circle. Key point: I said I could have used work-energy for the stopping part. I could NOT have used work energy for this turning part (well, I could use it but it wouldn't give me anything useful). There are two reasons why the work-energy principle won't do you any good. First, the speed of the car doesn't change during this motion. This means that there is no change in kinetic energy. Second, the frictional force is perpendicular to the direction of motion so that it does no work (we can discuss work done by static friction later).
Back to the turning calculation. I know an expression for the frictional force and I want the radius of the circle to be s. This gives:
And there you have it. If a car is traveling at a certain speed, it can stop in half the distance that it would take to turn.
I kind of like this result. Long ago, I took a driving class. You know, to learn how to drive. One think stuck in my mind. While driving, something came out in the road in front of me (I can't remember what it was). I reacted by swerving just a little into the next lane. The driving instructor used that annoying brake on the passenger side (that he would sometimes use just to show he was in control - I was going to stop, but he didn't give me a chance). Anyway, he said "always stay in your lane". He probably said that because he was so wise in physics even though he did smell funny.
Oh, it is probably a good idea to stay in your lane not only for physics reasons but also because you don't want to hit the car next to you (unless you are playing Grand Theft Auto - then that is encouraged).
I wonder if you could stop in even shorter distance? Is stopping the best way? Is there some combination of stopping and turning that could work?
Let me try the following. What if the car brakes for the first half and then turns for the second half. Would it hit the wall? First, how fast would it be going after braking for s/2 distance? The acceleration would be the same as before:
Using the same expression for the stopping distance from above, I get:
And this makes sense. If the car is stopping just half the distance, then it should have half the kinetic energy (which is proportional to v2). Ok, so if that is the new speed, what radius of a circle would it be able to move in? Again, using the expression from above:
Using this with half the distance - the total distance it would take to stop would be:
This is still greater than the stopping distance for just braking (which is s). But, did I prove that just stopping is the shortest distance? No. Maybe I just convinced myself to stop for now.
Here is a short bonus. Let me show that the work-energy principle is the same as that kinematic equation I was using. So, a car is stopping with just friction. The work done on the car by friction (and I can do this if I consider the car to be a point particle):
The work-energy principle says this will be the same as the change in kinetic energy of the car. If the car starts at a speed of v0 and stops at rest then:
See. Same thing.
How wide would the wall have to be so that it wouldn't matter if you brake or turn? Either way you would miss?
One of the things I remember being drilled into me during driver training was brake and avoid.
There's an interesting dynamics aspect to this, in that braking results in a transfer of the car's weight forwards, which gives you more traction to make the turn. In the absence of ABS, slamming on the brakes will result in your wheels locking up and thus a complete inability to steer (not to mention, an increased stopping distance).
What about hard left, then brake? After all, in America, that would get the driver's side away from the wall, so that if your stopping distance is too much, you at least avoid a head on collision, and provided you have no passenger, there shouldn't be serious injury. Because if s> the distance between the car and the wall, you really need to think in terms of minimizing damage.
For that matter, if you hit the wall at an angle instead of head-on, you'll slide along it instead of stopping instantly. This should reduce the deceleration forces acting on you, decreasing your odds of injury. So it seems to me that combined steering and braking is the way to go.
Don't forget the crumple zone you have if you hit front on
In Sweden you learn to break and swerve. Reason being, your normal obstacle (a moose, a human) is not all that wide, and the road is pretty likely to be empty apart from yourself. If there's lots of traffic, it's unlikely that there'd be a static obstacle on it after all.
There's a secondary consideration too. If you just swerve and fail, then you're likely to hit the object at close to your original speed. And as speed is by far the most important determinant of injury and death, that's a Bad Thing.
On the other hand, obstacles are very rarely really wide. Anything remotely likely to show up on a street is pretty narrow - a moose, a car or something like it. So once you're slowed down a bit chances are still pretty good that you'll be able to avoid it. And if you don't, you'll still hit at much lower speed than if you just swerved and failed.
Turn and then brake? Why didn't I think about that? I did brake then turn. Hmmm.. I will have to do a follow up post.
if you have a choice of a head on collision with, say an oncoming car, or swerving a bit so it isn't head on, choose the swerve option.
your momentum can be broken up into components parallel and perpendicular to the other objects motion. by avoiding headon, you can decrease the parallel component by the cosine of the angle, and thus decrease the magnitude of your change in momentum, the acceleration and thus the damage.
If your wall is infinitely wide (as stated in the original problem), you can show that braking is always the best option if you want to avoid a collision with the greatest car-wall separation (assuming the car is a point).
Just decompose your velocity vector into the component normal to the wall and a component along the wall.
If you turn (and avoid the wall) or do some combination of braking and turning:
1) the component normal must decelerate to zero at the point of closest approach
2) the component along the wall will increase to something nonzero.
If you just brake, you still have to do #1, but you avoid any forces required to achieve #2. Thus braking requires less total force.
So, for the infinite wall (and a point-mass car) braking along a straight line is superior to any braking/turning combination.
For the finite wall (or moose, as discussed above) it would probably require some algebra.
I agree with break and trun. With that, how fast would your speed have to be to flip your vehicle during your turn (depending on weight, center of gravity, width of tires, etc.)?
A lane change can be done in significantly less distance than either of the two extant options, but that will not work in the classic homework version of this problem where the road is completely blocked but there is a side street open if you turn. This works because you don't turn 90 degrees. Not even close. However, see my final observation below.
Minor detail in your solution: The standard problem assumes the same coefficient of friction for braking and turning, but this is only an approximation. Most tires have more grip under braking than when turning, which favors the braking solution even more.
What Anonymous Coward says is a result of what is known as the "friction circle" in racing. You trade sideways acceleration for fore-aft acceleration, which is why you brake and then turn ... leaving out all sorts of fine details. (Tires are complicated. Kinetic friction is a bit more than static friction as long as the slip is a small fraction of the velocity, then falls off rapidly once the tire is not rolling at all. This applies to both forward motion and the "slip angle" when cornering and is significantly different, depending on tire design, when cornering. The "friction circle" is usually an ellipse with the coefficient bigger for braking as noted above.)
The coefficient of friction when rolling, or (even better) rolling and slipping slightly, is bigger than when the tire is sliding sideways. This means that tossing the car so it slides sideways is a really bad solution and will result is significantly larger values for the variable "s" in your solution.
All of this ignores vehicle dynamics and traffic. If you are in the right car and are fully aware of the space around you, a lane change is the best way to avoid a problem in the road ahead. If you are not fully aware of the space around your vehicle, changing lanes can make things worse, and even skilled drivers are not always paying full attention to the road. Hence the driving instructor advice to stay in your lane. Worse, you might be in a van or SUV where sudden turning movements, particularly lane changes, with under-inflated tires can cause the vehicle to roll. Even passenger cars can get in trouble this way if one tire is off the pavement, which is why you should have been taught to slow down in a straight line before pulling the car back on the pavement from an unpaved shoulder.
@comment 1- You can still steer when braking, independent of whether or no the car has ABS. It's called the Emergency Brake, and in most vehicles it is attached only to the rear wheels.
I see I've been beaten to the Brake and Turn. I have done just that in a car- E-brake and a turn so that I do a complete 180. That would seem the best choice- and the distance needed to stop is far less than braking alone.
@11 - Unless you're a trained stunt driver, I wouldn't advise yanking the hand brake and then turning, because you're going to end up locking up the rear wheels and then going into an uncontrolled skid/slide.
But sure, it will look really cool.
@11 - The only advantage to doing a 180 is that you hit the wall going backwards. Your stopping distance will be greater because the coefficient of friction is less than with threshold braking or modern ABS while going straight.
The disadvantage might be that you kill the kids in the backseat because cars are not designed to crash that way.
Sure would make a great Mythbusters episode, however.