A car traveling at 53 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 65 cm (with respect to the road) while being brought to rest by an air bag. What magnitude of foce (assume constant) acts on the passengers's upper torso, which is 41 kg?
I changed 65 cm into .00065 km. I think i need to use F=ma so I need to find a. To find a I used d =1/2 (v0+v)t to find time and I got 2.45X10^-5. then I used a = v/t and i got 2163265.31 as a. Then i pluged m and a into F=ma and I got 88693877.71 for the answer. Is this correct? I wouldn't think the force would be this high.
Why not use
force*distance= 1/2 mass*velocity^2
That is the work done on the bag is equal to the change in KE of the passenger.
You should get the same answer as your technique, but I suspect the teacher was exploring the energy relationships in motion, as I laid out.
For information:
force*distance= 1/2 mass*velocity^2
force*distance= 1/2 mass*(avgveloctity*2)^2
force*distance= 2 mass*(avgveloctity )^2
force*avgvelociy*time= 2 mass*(avgveloctity )^2
force = 2 mass*(avgveloctity )/time
force = mass*(2avgveloctity )/time
force= mass*acceleration as you started from.
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