What is the magnitude of the average force that the air bag exerts on the driver?

[squintyeyes]

A 1900 kg car moving at 20.0 m/s crashes into a brick wall and is stopped. The driver's mass is 70.0 kg, and although the driver is not wearing his seatbelt, the air bag thankfully slows him to a stop in 0.100 s. What is the magnitude of the average force that the air bag exerts on the driver?
______________

(Side Note: Air bags and seat belts act to slow you to a rest in a longer time. By increasing the time, the force on your body decreases. An average force exceeding roughly 2.00 x 105 N will probably kill you.)

How would you do this problem?

 

[pgardn]

A 1900 kg car moving at 20.0 m/s crashes into a brick wall and is stopped. The driver's mass is 70.0 kg, and although the driver is not wearing his seatbelt, the air bag thankfully slows him to a stop in 0.100 s. What is the magnitude of the average force that the air bag exerts on the driver?
______________

(Side Note: Air bags and seat belts act to slow you to a rest in a longer time. By increasing the time, the force on your body decreases. An average force exceeding roughly 2.00 x 105 N will probably kill you.)

How would you do this problem?

Have you done impulse and change in momentum? The driver has an impulse supplied by the air bag that causes him to change momentum.

 

[squintyeyes]

so what would the equation look like? how would you solve it?

 

[pgardn]

so what would the equation look like? how would you solve it?

Have you studied F*t...?

 

[squintyeyes]

So i read up on this and did this. It was marked wrong can someone help me figure out why?

m*v = f*t
(1900 + 70)(20) = F * (0.1)
39400= 0.1F
F=394000

 

[pgardn]

So i read up on this and did this. It was marked wrong can someone help me figure out why?

m*v = f*t
(1900 + 70)(20) = F * (0.1)
39400= 0.1F
F=394000

I believe they were asking about the force on the driver (70 kg) and he was moving at 20 m/s and then 0 m/s (stopped). So the driver's change in momentum is due to a force acting over a 0.1 second time period.

 

[squintyeyes]

so would it be like this then?

m*v = f*t
(70)(20) = F * (0.1)
1400= 0.1F
F=14000

 

[pgardn]

so would it be like this then?

m*v = f*t
(70)(20) = F * (0.1)
1400= 0.1F
F=14000

yeps...

I am assuming on the left hand side of your equation you had final momentum being zero. So really you end up with a force that is (-). Which basically means the driver was accelerated in the opposite direction as his initial motion.