Seat-belt effects on passengers during a crash

[Alfos001]

Homework Statement

Hey, so I have found myself at a loss in trying to solve the forces that act on car passengers wearing their seatbelt vs a passenger without one when experiencing a crash.

I have tried approaching the problem in a few ways but cant find a way that is correct or makes sense and would like help in knowing how to solve it.

Relevant Equations

F=m*a
a=((vf^2-vi^2)/2(xf-xi))

I have attempted solving this problem several times on an long since finished online quiz. However no answer I gave seemed to be correct despite what I could reasonably understand regarding the effect of seatbelts and their effect on reducing force by extending the distance of deceleration. Although I can't retake this online exam anymore, I want to learn for the sake of my own peace of mind how to gauge the effects of seatbelts on passengers during sudden deceleration.

For my first attempt I added the respective distance over which each passenger decelerated.
The first passenger wearing his seat-belt experienced a total displacement (xf) of 0.15m + 0.3m =0.45m .
And the second passenger without their seatbelt had a total displacement (xf) of: 0.15m + 0.05m =0.20m .

Following that I converted the initial velocity (vi)= 30km/hr to (vi)= 8.333r m/s
And with that done I calculated the two forces as shown below:

F= m * [(vf^2-vi^2)/(2*(xf-xi))]

F1= 70* [(0^2 - 8.333^2)/(2*(0.45-0))] = 70* [-69.439/0.9] = -5,401N =-5.4kN

F2= 70* [(0^2 - 8.333^2)/(2*(0.20-0))] = 70* [-69.439/0.4] =-12,152N = -12kN

Submiting these as my answers however was marked as incorrect, I tried resubmitting the loads as positive forces instead on the off chance that I'd misassumed the direction of the loads but I was still just as wrong.
This lead lead me to question whether instead I was supposed to ignore the deceleration distance of the car (Which had been given and used to solve an earlier question in the quiz.
So by testing this alternative approach I naturally calculated much larger forces:


F1= 70* [(0^2 - 8.333^2)/(2*(0.3-0))] = 70* [-69.439/0.6] = 8,101N = -8.1kN

F2= 70* [(0^2 - 8.333^2)/(2*(0.05-0))] = 70* [-69.439/0.1] =48,607N = -49kN

The answers I got were also marked as wrong and the larger force value calculated by discounting the deceleration distance of the car also makes me suspect that this alternate approach to be incorrect.
So now I am left unsure of what else to try to calculate the forces that'd be acting on the passengers due to a crash. I feel as if I am missing something obvious in what I'm meant to do to work out the force acting on a passenger but nothing's coming to mind and I would really appreciate helpful pointers to put my brain thinking in the right direction.

Also I hope that my layout of the problem and methods I've used is alright and I've explained my problem well enough. If not I will try to give clearer information if necessary. Any help I can get will be appreciated, thank you.

 

[hutchphd]

The actual accident would give the best case for the belted passenger and the worst case for the unbelted one: the belted passenger will decelerate over the entire distance traveled but the unbelted one will travel the 30 cm to the dashboard and hit the now stationary dashboard at full speed. splat. Been there, done that...

 

[Tom.G]

Please take into account that an un-restrained passenger decelerates only when they hit something, such as a dashboard or door.

Ahh, hutchphd was 30 seconds quicker!

 

[haruspex]

Homework Statement:: Hey, so I have found myself at a loss in trying to solve the forces that act on car passengers wearing their seatbelt vs a passenger without one when experiencing a crash.

I have tried approaching the problem in a few ways but can't find a way that is correct or makes sense and would like help in knowing how to solve it.
Relevant Equations:: F=m*a
a=((vf^2-vi^2)/2(xf-xi))

I have attempted solving this problem several times on an long since finished online quiz. However no answer I gave seemed to be correct despite what I could reasonably understand regarding the effect of seatbelts and their effect on reducing force by extending the distance of deceleration. Although I can't retake this online exam anymore, I want to learn for the sake of my own peace of mind how to gauge the effects of seatbelts on passengers during sudden deceleration.

View attachment 264388

For my first attempt I added the respective distance over which each passenger decelerated.
The first passenger wearing his seat-belt experienced a total displacement (xf) of 0.15m + 0.3m =0.45m .
And the second passenger without their seatbelt had a total displacement (xf) of: 0.15m + 0.05m =0.20m .

Following that I converted the initial velocity (vi)= 30km/hr to (vi)= 8.333r m/s
And with that done I calculated the two forces as shown below:

F= m * [(vf^2-vi^2)/(2*(xf-xi))]

F1= 70* [(0^2 - 8.333^2)/(2*(0.45-0))] = 70* [-69.439/0.9] = -5,401N =-5.4kN

F2= 70* [(0^2 - 8.333^2)/(2*(0.20-0))] = 70* [-69.439/0.4] =-12,152N = -12kN

Submiting these as my answers however was marked as incorrect, I tried resubmitting the loads as positive forces instead on the off chance that I'd misassumed the direction of the loads but I was still just as wrong.
This lead lead me to question whether instead I was supposed to ignore the deceleration distance of the car (Which had been given and used to solve an earlier question in the quiz.
So by testing this alternative approach I naturally calculated much larger forces:


F1= 70* [(0^2 - 8.333^2)/(2*(0.3-0))] = 70* [-69.439/0.6] = 8,101N = -8.1kN

F2= 70* [(0^2 - 8.333^2)/(2*(0.05-0))] = 70* [-69.439/0.1] =48,607N = -49kN

The answers I got were also marked as wrong and the larger force value calculated by discounting the deceleration distance of the car also makes me suspect that this alternate approach to be incorrect.
So now I am left unsure of what else to try to calculate the forces that'd be acting on the passengers due to a crash. I feel as if I am missing something obvious in what I'm meant to do to work out the force acting on a passenger but nothing's coming to mind and I would really appreciate helpful pointers to put my brain thinking in the right direction.

Also I hope that my layout of the problem and methods I've used is alright and I've explained my problem well enough. If not I will try to give clearer information if necessary. Any help I can get will be appreciated, thank you.

I would say your second attempt is the expected approach. Since the stopping distances are given relative to the ground, the car's stopping distance is irrelevant.

Did you try +8.1 and +49? It asks for magnitudes.

Otherwise, the caveat regarding precision makes me suspect this may be where you are not meeting expectation. For the seatbelt case, you have two significant figures for each datum, so a suitable answer would be 8.1kN, but for the unrestrained case the 5cm is only one sig fig, so round to 5E1 kN (or whatever you use for scientific notation).

But I was careful to write "expected approach". The whole basis of the question is a classic blunder. The definition of average force is change in momentum divided by elapsed time, and don't let anyone tell you otherwise. You need to know the elapsed time, which can be different with different force profiles. The KE method assumes a constant acceleration.
Some try to defend it as "average over distance", but then you have to accept the notion of average acceleration as being an average over distance. It is easy to construct examples to demonstrate the crazy answers that can give.

 

[berkeman]

Been there, done that...

Unbelted? Bad, bad @hutchphd -- glad you survived, though!

 

[hutchphd]

Thanks. Yes I was 20 and, of course, not subject to physical laws. But it turns out you do not want to be in a VW beetle hit head on by a drunk Chevy impala. A tough lesson and I was fortunate to survive. It was my last unbelted auto ride!

 

[berkeman]

It was my last unbelted auto ride!

Glad you survived! Yeah, we all probably have similar lessons learned. Mine was with a rogue wave when climbing out of the ocean up the rocks when abalone diving. I was much more observant after that...

https://en.wikipedia.org/wiki/What_does_not_kill_me_makes_me_stronger

 

[Alfos001]

Thanks everybody for the feedback, I'm at least relieved to know that my approach to this problem wasn't too far wrong than what I'd dreaded it to be. At least with the second attempt anyway. I think I assumed the stopping distance needed to include both the car's and the seatbelt's only because 8.1kN struck me as a dangerously heavy load for a seat-belt wearer to experience which made me question if there were meant to be other factors that'd make the safety equipment seem more effective.

I would say your second attempt is the expected approach. Since the stopping distances are given relative to the ground, the car's stopping distance is irrelevant.

Did you try +8.1 and +49? It asks for magnitudes.

Otherwise, the caveat regarding precision makes me suspect this may be where you are not meeting expectation. For the seatbelt case, you have two significant figures for each datum, so a suitable answer would be 8.1kN, but for the unrestrained case the 5cm is only one sig fig, so round to 5E1 kN (or whatever you use for scientific notation).

But I was careful to write "expected approach". The whole basis of the question is a classic blunder. The definition of average force is change in momentum divided by elapsed time, and don't let anyone tell you otherwise. You need to know the force profile. The KE method assumes a constant acceleration.
Some try to defend it as "average over distance", but then you have to accept the notion of average acceleration as being an average over distance. It is easy to construct examples to demonstrate the crazy answers that can give.

I am fairly certain that I tried both positive and negative forms of the magnitudes but it is possible that in my efforts to try a different approach I neglected to try both forms for both approaches.
Precision could also be where I've stumbled as that's caught me out a few times before.

Incidentally this question was part of a quiz for momentum, collisions and impulses. The crash was caused by an inelastic collision between cars of equal mass & velocity traveling in opposite directions, which I think would have resulted in the final momentum cancelling out to 0m/s (Which is perhaps also why the force on the seatbelt wearer appears as lethal as it does for the non belt wearer).
There was also no information regarding the contact time of the collision to use for impulse calculations to be possible either.
Thus the only approach left has been to calculate acceleration over distance traveled so it's good to know my approach wasn't too far off.

Also I'm glad to learn that your experience in a car without a seatbelt wasn't so fatal either, Hutchpad. :)