Physics behind Car Safety

Everywhere around the world, cars are used throughout in most people’s everyday life. From just going to the super market across the road, and going across the country boarder, cars are one of the most essential sources of item, to live a typical life style in developed countries. And although cars seem to look very simple, and can be used very simply, it can be said to be one of the most complicated machines in the world. And to get to the design of cars we use now, billions of hours were spent on research.

And most of these hours have been spent just on safety issues. Imagine you’re in a car, and a baby crawled out into the road 30 metres ahead of the car. This is not a major issue, as all you have to do is step on the break, and the car will stop before it hits the baby. Believe it or not, 100 years ago, if this happened, the only way for you to avoid the crash was to turn your steeling wheel, as the technology in today’s society did not exist 100 years ago. Also, the roads itself have been designed so sympathetically, to insure safeties for not only for drivers and passengers, but also for pedestrians. Road rules also exist to lessen the amount of miserable accidents.

Every year, or even every month, cars are getting safety and even more safe, and roads keep advancing, for example, the roads in the Japanese highway are made out of concrete which absorbs water, which means, there is less slipping and sliding of the cars. All these features do not exist just to save our own lives, but also to save everyone around us in our living community. And this car safety, can be investigated thoroughly through the use of Physics, and throughout the report, the wide world of physics behind car safety will be investigated.

Vehicle DesignMany features found in cars are there to reduce the force component of the impulse, or the change in momentum. By reducing the force component of the impulse, the impact on the people inside the car will have less effect. This can be shown through the formula of impulse. △p = F x t, where △p is the change in momentum or the impulse, F is force in Newtons and t is time in seconds.

Examples of safety features that use impulse in cars are, seat belts, air
bags and crumple zones.

1. Seat Belts (Impulse/Momentum)A seat belt is one of the most imperative safety features. The main idea of the seat belt is to apply a force called the ‘stopping force’ to the body over a long period of time as possible, comparison to the instant shock of hitting the windscreen. As stated above, the formula for Impulse is △p = F x t. And due to the law of conservation of momentum (which states that, the momentum transferred from one object to another is the same), by increasing the time taken to stop, it means that smaller force is required to reach the same change in momentum. And therefore, smaller force means that less damage is done to the body.

Case:For example, lets assume a 60kg person drives a car into a wall at 50km/hr (13.9m/s).

To begin with, the change in momentum is pFinal – pInitial. And the formula for momentum is, p = m x v, where p is the momentum, m is mass and v is velocity.

Therefore,△p = pF -pI= mv – mu= 60 x 13.9 – 60 x 0= 834kgm/sTherefore, the change in momentum is 834kgm/s.

Without a seat belt, it is said that, the average time taken for the body to decelerate is 0.01 seconds. With a seat belt, the deceleration time is usually increased to about 0.5s.

Without a seat belt,△p = F x tF = △p/tF = 834/0.01F = 83400NTherefore, without a seat belt, the force exerted to the body when it hits the windscreen is, 83400N.

With a seat belt,△p = F x tF = △p/tF = 834/0.5F = 1668NTherefore, with a seat belt, the force exerted to the body is, 1668N.

As it can be clearly be seen above, by having a seat belt on, and by having nothing on, there is a significant difference. If this person did not have a seat belt on, and ended up with 83400N of force exerted, this is enough force to simply crush a skull. 83400N is equivalent to, 8 adult elephants standing on you. If this person did have a seat belt on, the chance is that, this person will have 1668N(equivalent to 1 sumo wrestler standing on you) of force exerted to the person’s body, where the seat belt is touching, which could leave that person with major bruises, which is certainly better than a crushed skull.

So one of the most ideal methods to maximize car safety is to improve on the seatbelt’s stretching time. As you can see above, just 0.1 seconds of extra stretching time can minimize the force exerted to the body by extraordinarily more less, and therefore, it can be said to be one of the most important safety feature in car safety.

Road DesignIf a car is to drive on a circular path, there are two types of ways. The first method is using circular motion (and friction) in order to turn. The second method is banking the circular path, in order to use the weight component to help the car turn.

2. Cornering using Circular Motion and FrictionIf you are driving and you are to turn on a circular path, cornering is depended by circular motion or “centripetal acceleration” by the using the friction of the tyres. To find out the force supplied by the tyres on the road, the formula for circular motion can be used.

Fc = mv2/rAnd this formula can be fixed around to make calculations easier by substituting the formula for Friction, Ff = μFn.

Ff = FcμFn = mv2/rμmg = mv2/rμg = v2/rr = v2/μgr = v2/μgAs it can be seen above, when it comes to cornering, mass is an irrelative variable, so therefore, when designing a road, the only 2 essential things you have to worry about is, the radius of the circle path, and the speed of the car.

Case:Assume a car is trying to turn left in a speed of 40km/hr, to head towards perfectly right angle to the highway. What radius would the circular path need to have, in order for the car to turn without problem?r = ?v = 40km/hr = 11.1m/secμ = 0.7 (average frictional coefficient on highways in Australia between the road and the tyre)g = 9.8m/s/sr = v2/μgr = 11.12/ 0.7×9.8r = 123.46/6.86r = 18 metresTherefore, for this car to turn left, right angle to the highway in the speed of 40km/hr, the circular path needs to have a radius of at least 18m, in order for the car to turn without any problem.

3. Banking of Circular Path (Banking)By banking corners, it allows cars to turn much more essentially, as banking allows the weight component to help the car turn. If we bank a corner, there are components of the normal force (mgsinø), which is in the direction right angle to the path. If the corner has a banking angle of ø, then the force component towards the centre will be,Fnsinø = mv2/rThe normal force, Fn can be found by replaced by the formula, Fn =mg/cosø.

Therefore,Fnsinø = mv2/rsinø x mg/cosø = mv2/rmg x sinø/cosø = mv2/rmg x tanø = mv2/r (sinø/cosø = tanø)g x tanø = v2/rtanø = v2/rgTanø = v2/rgCase:Assume a car is trying to turn a corner with a radius of 15m, in a speed of 40km/hr. In the last feature, 2.Cornering using Circular motion and friction, a car with a speed of 40km could not turn a corner with a radius of 18m or less. But by banking, it is physically and realistically possible. Lets see.

r = 15mv = 40km/hr = 11.1m/secg = 9.8m/s/sø = ?degreestanø = 11.12/ 15×9.8tanø = 123.46/147tanø = 0.74ø = tan-1(0.74)ø = 36degreesTherefore, for this car to turn a corner with radius of 15m in a speed of 40km/hr, the road must be banked in an angle of 36degrees from the ground.

Road Rules4. Traffics (Kinematics)Everywhere around the world, there are always signs in order to show signals to tell the drivers what to do, to insure safety for the driver, passengers and for pedestrians. One example of a sign is the “Traffic Light”. They show the driver, when they should step on the break. As everyone knows, when the traffic lights are green, it shows that they can keep driving through, when it is on yellow, it warns us that the light is going to change red in about 3 seconds, so the driver can step on the break. But, if a car is 60 metres away from the intersection straight after seeing the yellow light flashing, is the driver meant to stop or keep going? If a car is 100 metres after the yellow light flashing, is the driver meant to stop or keep going? In this section, I would like to make a road rule on when to step on the breaks after seeing the yellow light flashing.

Case:To begin with, we will assume that a car is driving in 50km/hr, and the car has a deceleration rate of 7.8m/s/s (average deceleration rate of a four wheel drive). The formula we will be using is,s = ut + 1/2at2Starting off, we must find out how far the car can travel in the 3-second period (the time for the yellow light to turn red).

u = 50km/hr =13.9m/sa = 7.8m/s/st = 3secondss = ?s = ut + ½at2s = 13.9 x 3 + ½ x 7.8 x 32s = 41.7 + 35.1s = 76.8mSo after all, the calculations clearly shows that, if the driver sees the yellow light flashing, less than 76.8m away from the intersection, there is no need to stop, and therefore, there should be no problem keep going. On the other hand, if the car is more than 76.8m away from the intersection, the driver should be ready to step on the breaks, to avoid car crashes.

Note: We must remember that, although you are less than 76.8 away from the intersection and there is no need to stop, it is always better stopping early, and must realise that, 76.8m is only the “boarder line” length, and this boarder line length cannot insure 100% safety.

5. Saving Innocent Pedestrians (Kinematics)If you are driving, and all of the sudden a children jumps out 20m in front of the car, what would you do? A normal person will step on the break to avoid running over the children. But if a person is 10m in front of the car, is there any point of stepping onto the break? Will the car stop before it hits the person? In this section, I would like to find out the distance for a car to stop.

The formula, which we will be using, is,v2 = u2 + 2asCase:Assume a car is driving in 50km/hr, and suddenly, the driver sees a child jumping out into the road, 15m in front of the car. Assume the car has a deceleration rate of 7.8m/s/s.

Now we must find, how far the car will need in order for the car to come to rest from 50km/hr.

v = 0km/hr = 0m/su = 50km/hr =13.9m/sa = -7.8m/s/ss = ?v2 = u2 + 2as02 = 13.92 + 2 x -7.8s0 = 193.21 -12s15.6s = 193.21s = 12.39mTherefore, for this car to come to rest, it needs approximately 12m before it comes to rest. Therefore, in this case, the driver should not worry about anything else but to step on the break.

On the other hand, if a child jumped out 10m in front of the car, the only way for the driver to save that child is to move its steering wheel and avoid the child.

Note: The driver must remember, he will not step on the breaks, as breaks can cause drift, which can lead to slamming the child with huge momentum with the side of the car.

ConclusionIn this world, there are many safety features present in cars and on roads. These features may or may not save our life according to the type of accident that happened. Vehicles become more and more improved in the safetiness factor, and therefore, allow drivers and us pedestrians to live in our daily community with comfort and safety. Physics is what helps this improvement, and through this assignment, the wide world of car safety was able to be touched. Although it was only a very small particle of the enormous world of physics, we must understand that this was an incredibly important big first step.

Bibliography

SearchEngine:www.google.com.auwww.askjeeves.com.auWebsites:http://www.reachoutmichigan.org/funexperiments/agesubject/lessons/newton/airbags.html (online) (accessed: 19th July 2006)http://www.tpt.org/newtons/9/airbags.html (online) (accessed: 19th July