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.

Step by step solution

01

Given

Mass of person, m =60kg

Initial speed of car, u = 15m/s

Distance with seat belt,$s_1$=1m

Distance without seatbelt,$s_2$=5mm=0.005m

Finial speed, v = 0

02

Explanation

a.

Using $V^2$=$u^2$+2as,

0 = ${\left(15\right)}^2$+2a(1)

a = $\frac{-225}{2}$= -112.5$m/s^2$

Using Newton's second law. F = m.a

$F_1$= (60)(112.5) = 6750 N

Therefore, net force on the person is 6750 N

03

Explanation 

b.

Using $V^2$=$u^2$+2as,

0 = ${\left(15\right)}^2$+2a(0.005)

a = $\frac{-225}{0.01}$= -22500$m/s^2$

Using newton's second law, F = ma,

$F_2$= (60)(22500) = 1350000 N

The net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag is 1350000 N.

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