Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. cars are designed with a "crumple zone" in the front of the car. in the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. an occupant restrained by seat belts and air bags decelerates with the car. by contrast, an unrestrained occupant keeps moving forward with no loss of speed (newton's first law! ) until hitting the dashboard or windshield. these are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about 5 mm .
a. a 60 kg person is in a head-on collision. the car's speed at impact is 15 m/s. estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys. b. estimate the net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag. c. how does these two forces compare to the person's weight?

a) -6,750 N  b) -1.35*10⁶ N  c) 11.5 times d) 2,295 times

Explanation:

a) In order to answer the question, we can apply Newton's 2nd Law to the person, as follows:

Fnet = m*a

Assuming that the deceleration is uniform, we can use any of the kinematic equations.

In this particular case, examining the givens that we have (the final speed (which is 0), the initial speed (impact speed), and the distance of deceleration (1 m) the most useful equation is the following:

vf²-v₀² = 2*a*d

Replacing by the givens, and solving for a, we get:

a = (15m/s)² / 2* (1 m) = -112.5 m/s²

With this value of a, we can get the net force F:

F = 60 kg* (-112.5 m/s²) = -6,750 N

b) For this part, same reasoning applies, the only difference being the deceleration distance, which for this case is only 5 mm.

We can apply the same kinematic equation:

vf²-v₀² = 2*a*d

Once again, replacing by the givens, and solving for a, we get:

a = (15m/s)² / 2* (0.005 m) = -22,500 m/s²

With this value of a, we can get the net force F:

F = 60 kg* (-22,500 m/s²) = -1.35*10⁶ N

c) If we compare the forces that  we got above with the weight of the person (which is the same to compare the acceleration with g), we have, for the first case (person restrained) a value of  approximately 12 g, but for the unrestrained case, we got a value of 2,295 g!

your answer is c! hope this : )