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 1.00m 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 5.00mm .

1. A 73.0kg person is in a head-on collision. The car's speed at impact is 15.0m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

2. Estimate the net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag.

3. What is the force in part A in terms of the person's weight?

4. What is the force in part B in terms of the person's weight?

The answers to the question are;

1. The net force on the person if he or she is wearing a seat belt and if the air bag deploys is 8212.5 N.

2. The person if he or she is not restrained by a seat belt or air bag = 1642500 N

3. The force in part A in terms of the person's weight is

11.47 × the person's weight

4. The force in part B in terms of the person's weight is

2293.6 × the person's weight

Explanation:

To solve the question, we note the following equations of motion

v² = u² - 2×a×s

Where:

v = Final velocity of an object = Object at rest = 0 m/s

u = Initial velocity = 15.0 m/s

a = Acceleration

s = Distance

Over a stopping distance where the driver wears a seat belt and has an airbag, we have

s = 1.00 m

Therefore, v² = u² - 2×a×s gives

0² = (15.0 m/s)² - 2 × a × 1.00 m which gives

2 × a × 1.00 m = 225 m²/s² and

a = (225 m²/s²)/(2×1.00 m) =  112.5 m/s²

The net force then becomes

Force, F = Mass, m × Acceleration. a = 73.0 kg × 112.5 m/s² = 8212.5 kg·m/s²

The net force on the person if he or she is wearing a seat belt and if the air bag deploys =  8212.5 N

2. The net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag is given by when

s = 5.00 mm = 0.005 m

Therefore

v² = u² - 2×a×s gives

0² = (15.0 m/s)² - 2 × a × 0.005 m and

a = (225 m²/s²)/(2×0.005 m) = 22500 m/s² and

Force, F = Mass, m × Acceleration. a = 73.0 kg × 22500 m/s² = 1642500 kg·m/s² = 1642500 N

The person if he or she is not restrained by a seat belt or air bag = 1642500 N

3. The weight of the person can be found by

Weight = Mass × Gravity = 73.0 kg × 9.81 m/s² = 716.13 N

Dividing the net force acting on the person if he or she is wearing a seat belt and if the air bag deploys is given by 8212.5 N/716.13 N = 11.47

Therefore the force in part A in terms of the person's weight is

11.47 × the person's weight

4. Dividing the net force acting on the person if he or she is not restrained by a seat belt or air bag by the weight of the person, we have

1642500 N/716.13 N = 2293.6

Therefore the force in part B in terms of the person's weight is

2293.6 × the person's weight

unlike poles attract

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