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Turn or go straight? Quick!

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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.

Stopping

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:

i-449ed916b9f5d3d7aab2a658926046b9-2010-08-04_untitled.jpg

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.

i-1ea63c7b92047077c0a3735932136c5f-2010-08-04_la_te_xi_t_1_5.jpg

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:

i-755d05d3d5993bfe5fd7cb9c47c56883-2010-08-04_la_te_xi_t_1_6.jpg

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.

i-19a001f7275b44902b9ea84d6af00e85-2010-08-04_la_te_xi_t_1_7.jpg

There you have it. That is how far the car would need to stop. Quick check – does it have the right units? Yes.

Turning

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?

i-0517f942cf57cf09d76f7d918e755fbf-2010-08-04_untitled_1.jpg

For an object moving in a circle, the following is true:

i-32eb8215e6f7d0a651c0f5d532295ef7-2010-08-04_la_te_xi_t_1_10.jpg

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:

i-594f719b83fcea02151167844171de14-2010-08-05_la_te_xi_t_1_11.jpg

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).

Another question

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:

i-99554f7660b138dc17ff51d638bc0d6b-2010-08-05_la_te_xi_t_1_13.jpg

Using the same expression for the stopping distance from above, I get:

i-738aa7f4589988b81effeb2f256cacd8-2010-08-05_la_te_xi_t_1_14.jpg

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:

i-23371749f1471d3c2d7fe77ffe1d297f-2010-08-05_la_te_xi_t_1_15.jpg

Using this with half the distance – the total distance it would take to stop would be:

i-dd8bdcf1df712be21b66b92764caa3b0-2010-08-05_la_te_xi_t_1_16.jpg

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.

Bonus

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):

i-b3468a700f5e86b20453e677b14474a2-2010-08-05_la_te_xi_t_1_18.jpg

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:

i-6538455d42b6168b70023c0e1039d1a4-2010-08-05_la_te_xi_t_1_19.jpg

See. Same thing.

Homework

How wide would the wall have to be so that it wouldn’t matter if you brake or turn? Either way you would miss?


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