- #1

Alfos001

- 2

- 1

- 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.

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.