Figure 1 illustrates an unbelted person in the front passenger’s seat. Notice th

Question

Figure 1 illustrates an unbelted person in the front passenger’s seat. Notice that there are about two feet between him and the windshield. In a frontal car crash, he will continue at the pre-crash speed of the car until he hits the dashboard and/or windshield. If we assume that there is about ½ inch of flexibility in the windshield and dashboard, that means he has only ½ inch of space in which to experience this deceleration, resulting in his unprotected body being subjected to very large and dangerous forces.

   Figure 2 illustrates the same person in the same seat, protected by a seat belt + air bag system. This means that the same two-foot space is now used to help him decelerate over a much larger distance.

2. C. Calculate the acceleration, in feet per second, necessary to go from a velocity of 30 miles per hour to zero miles per hour in a distance of ½ inch. Show your work. You can type this in, or hand write it, scan it and insert it here.

Hints: 30 miles per hour is 44 feet per second. The equation needed to solve this problem is in Section 2.4 of the Interactive Textbook. The problem has not told you anything about time, because time is not needed to solve the problem.

2. D. 1. Calculate the acceleration, in feet per second, necessary to go from a velocity of 30 miles per hour to zero miles per hour in a distance of 2 feet.

2. D. 2. Show your work.

2. E. The acceleration due to gravity is 32 feet/second2. This is called “one G”.

   1. How many G’s did the passenger experience, unbelted with no air bag? _______________

   2. How many G’s did the passenger experience, with the seat belt + air bag system? _______

        (Note: This is an estimate. G forces in an actual accident may be greater.)

2. F. Briefly comment on why this protection from sudden deceleration makes the seat belt + air bag system an effective method for preventing injuries in car crashes.

Explanation / Answer

2. C.

given

vi = 44 ft/s

vf = 0 ft/s

d = 0.5 in = 0.041666 ft

deceleration = a

hence

2*d*a = vi^2 - vf^2

a = 23232 ft/s/s

D. here, d = 2ft

hence

a = 484 ft//s/s

E. g = 32 ft/s

so in case 1, a = 726g

case, 2 a = 15.125g

3. the body can take 15.125g as compared to 726g and hence this is an efficirent system to reduce impact acceleration.

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