How Do You Calculate the Average Force on a Driver in a Crash?

[jayberd]

Problem:
A race car skidding out of control manages to slow down to 90km/h before crashing head-on into a brick wall. Fortunately, the driver is wearing a safety harness. Using reasonable values for the mass of the driver and the stopping distance, estimate the average force exerted on the driver by the safety harness, including its direction. Neglect any effects of frictional forces on the driver by the seat.

Equations:
F=ma this is net force, but what is average force?
v_x=v_ox+a_xt


I tried using the v_x equation to find the acceleration. I guessed the initial velocity of the car/driver was 130km/h and I also guessed it would take 5 seconds to bring the velocity down to 90km/h. I ended up getting a really tiny force no matter how I tried it. I also just tried finding the acceleration the normal way, but the same thing. Anyone else have a good way of going about this?

 

[cepheid]

Welcome to PF jayberd!

The idea with the average force is that the safety harness applies a force to the driver over some distance (i.e. it allows him to move forward somewhat before stopping). In general, that force might vary over that distance in a complicated way (it won't be constant). We don't have enough information to say how the force that the safety harness applies varies with position during this stopping motion. All we can speak of is some sort of average force, where the average is taken over the distance interval.

The good news is that the total work done on the driver (which we DO know because we know his initial and final kinetic energies) will still be equal to that distance-averaged force times the total distance. So you don't have to actually take the average (which is good, because again, you don't have enough information to do so). Based on the total work done, you can work out what the average force had to have been.

We're talking about the deceleration that occurs from the moment of hitting the wall to the moment of the driver coming to a full stop. So 90 km/h is the INITIAL condition here, not the final one. The final condition is that the driver is at rest.

 

[jayberd]

Welcome to PF jayberd!

The idea with the average force is that the safety harness applies a force to the driver over some distance (i.e. it allows him to move forward somewhat before stopping). In general, that force might vary over that distance in a complicated way (it won't be constant). We don't have enough information to say how the force that the safety harness applies varies with position during this stopping motion. All we can speak of is some sort of average force, where the average is taken over the distance interval.

The good news is that the total work done on the driver (which we DO know because we know his initial and final kinetic energies) will still be equal to that distance-averaged force times the total distance. So you don't have to actually take the average (which is good, because again, you don't have enough information to do so). Based on the total work done, you can work out what the average force had to have been.

We're talking about the deceleration that occurs from the moment of hitting the wall to the moment of the driver coming to a full stop. So 90 km/h is the INITIAL condition here, not the final one. The final condition is that the driver is at rest.

so I end up having to solve for
0km/hr=90km/hr+a_xt
but where does t come from? s=dt? or can I use displacement in terms of velocity?

 

[cepheid]

so I end up having to solve for
0km/hr=90km/hr+a_xt
but where does t come from? s=dt? or can I use displacement in terms of velocity?

Umm, did you read my post? I think the best way of solving this problem is just to compute the total work done on the driver and equate it to the (average force)*(distance).

 

[cepheid]

Or you could use the kinematic formula for velocity in terms of displacement, the only difference from usual being that a is interpreted as being the average acceleration (with the average being as I described before).

 

[SammyS]

I suspect they're asking for the force exerted on the driver as he slows from 90km/h to 0km/h as the car crashes into the wall, crumpling up the front end of the car.

"... Using reasonable values for the mass of the driver and the stopping distance, estimate the average force exerted on the driver by the safety harness ..."

Reasonable mass & stopping distance might be ~75kg and ~1m respectively, both ± 30% or so.