Are Seatbelts Strong Enough to Protect Us During a Car Crash?

In summary, the conversation discusses an equation for seatbelts and how to calculate the impact force in a car crash. The correct formula for calculating acceleration is v^2 = v0^2 - 2ax. It is also mentioned that modern cars have collapsible front ends to reduce the deceleration force during a collision. The conversation concludes that seatbelts are important for absorbing energy and preventing serious injuries in car accidents.
  • #1
Dingoz
2
0
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Ive been trying to get my head around an equation for seatbelts. Hyperphysics has an auto calculator and I've been trying to do it algebraically and still cannot get the same answer.

Here is what i have done...

In relation to a car traveling at 50kmph with a driver which weighs 75kg and stops with in 1 foot.
Car Crash Example

m = 75 kg, initially traveling at 50kmph = 13.88888889m/s d= 30 cm
Distance time interval =
dt = 0.3 m / 13.88888889 m/s = 0.022 s

During this 220th’s of a second it is decelerated from 13.88888889 m/s to zero, so
a = dv/dt = 13.88888889 m/s / 0.022 s = 631 m/s^2

So the impact force is
F = m x a = 75 kg x 631 m/s^2 = 47348N ...

But it is completely different answer, i believe my acelleration calculation is wrong BUT cannot seem to find out how
 
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  • #2
Notice that the velocity over the 1 foot is not constant, so the time interval you calculated is incorrect. In fact, you should use the v^2 = v0^2 - 2ax formula to calculate the acceleration (assumed to be constant here), and then the force.

Hope that gives you the right answer.
 
  • #3
0 = 192.9 + 2*A*0.3

so do i...

0 =192.9 + 2*A*0.3 -2*A*0.3
- 2*A*0.3 = 192.9
-2*A*0.3/0.3 = 192.9/0.3
-2*A = 643

-2*A/-2*A = 643/-2*A
0 = 643 / -2*A/A
0/A = 643 /-2
A = 643/-2

A = 643 / -2
A = 321.5 ms^2 (- Negative due to deceleration... )
 
  • #4
A/g = 321.5/9.81 = ~33 g's. This deceleration would be pretty rough on anyone. This is why modern cars have collapsible front ends, so perhaps d -> 100 cm and the deceration is reduced to ~10 g's.
 
  • #5
Well, even so, it would be for hitting a wall, which would really hurt. Compared to hitting another car, the collision time is a lot shorter.
 
  • #6
Dropping dead on the a concrete floor will produce a lot more G's than that, and no drunkard die on me for just that. 33 G is quite reasonable as seatbelts absorb some energy before stopping you. The energy from a 50 Km/h collision should be like a drop from a 9 m height.

Seatbelts use the fact that you have a certain elasticity in your body, so broken bones or strained muscles are acceptable. Without them you would hit the dashboard which has little elasticity so much of the energy from the impact goes to your vitals (brain etc.)
 
  • #7
Yay seatbelts!

9 m = 3+ stories. Add to that the fact that your centre of mass will travel further than 0.3 m over the deceleration, the deceleration is less than 33 G's. Since a 3 storey fall is likely to break something or another, 33 G's is too much to get through without significant injury.
 
  • #8
jix said:
Yay seatbelts!

9 m = 3+ stories. Add to that the fact that your centre of mass will travel further than 0.3 m over the deceleration, the deceleration is less than 33 G's. Since a 3 storey fall is likely to break something or another, 33 G's is too much to get through without significant injury.

0.3 m is about right for a seatbelt but dropping on concrete or hitting the dashboard will decelerate you in only a few cm ( 2 - 5 maybe ) that means a 6-15 fold increase in G force experienced, and people still tend to survive 450 G's. There are very few fatal accidents at 50 Km/h that is why it's the European city speed limit.
 

1. What is the definition of "physics impulse"?

Physics impulse is a term used to describe the change in momentum of an object caused by an applied force over a period of time.

2. How is physics impulse calculated?

Physics impulse is calculated by multiplying the force applied to an object by the time period over which the force was applied.

3. How does physics impulse relate to Newton's laws of motion?

According to Newton's second law of motion, the change in an object's momentum is directly proportional to the applied force and inversely proportional to the time period over which the force was applied. This is the basis for the concept of physics impulse.

4. What are some real-life examples of physics impulse?

Some common examples of physics impulse include hitting a baseball with a bat, a soccer player kicking a ball, or a car coming to a sudden stop when brakes are applied.

5. How does the magnitude of force and time period affect physics impulse?

The greater the force applied to an object and the longer the time period over which the force is applied, the greater the change in momentum and therefore, the greater the physics impulse. This is why it is important to consider both force and time when studying physics impulse.

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