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

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The discussion centers on the physics of seatbelt effectiveness during car crashes, particularly at a speed of 50 km/h. Calculations reveal that deceleration can reach approximately 33 G's, which is significant but manageable due to the energy absorption capabilities of seatbelts. The conversation highlights that modern car designs, including collapsible front ends, help reduce the impact forces experienced by occupants. Comparisons are made to the forces experienced in falls from heights, emphasizing that seatbelts mitigate severe injuries by allowing for some elasticity in the human body. Overall, while the forces involved in a crash can be high, seatbelts play a crucial role in enhancing passenger safety.
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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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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.
 
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... )
 
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.
 
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.
 
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.)
 
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.
 
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.
 

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