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

Question

What is the force in part A in terms of the person's weight? What is the force in part B in terms of the person's weight? Express your answer using two significant figures. What is the force in part B in terms of the person's weight? Express your answer using two significant figures. 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 1m 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 What is the force in part A in terms of the person's weight? Express your answer using two significant figures. surfaces, and the unfortunate occupant then decelerates over a distance of only about 5mm . What is the force in part B in terms of the person's weight? Express your answer using two significant figures.

Explanation / Answer

a)because car is travelling at 10m/s.. and the car crumples 1m, that means the accident took 0.1seconds.

So the force would be:

Impulse = change in momentum

F.t = m(change in v)

F = 65kg(0 - 10m/s) / 0.1second

... = 6500Newtons

b) without restraint and seat belts, the person is still travelling at 10m/s until he/she hits the dash board... so over 5mm the time duration would be 0.005/10 = 0.0005seconds

So using the same eq.

Impulse = change in momentum

F.t = m(change in v)

F = 65kg(0 - 10m/s) / 0.0005second

.. = 1300000N = 1300kN

c) the person's weight = mg = 65 x 9.8 ~ 650N

So with the airbag/seatbelt, he/she would feel the force of ~ 10x his/her weight

d) 1300 000N ~ to 650N about a factor of 2000 times.

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