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Q. 42

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 156

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Short Answer

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 1mas 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 60kg person is in a head-on collision. The car’s speed at impact is 15m/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

(a) The net force on the person if he/she is wearing a seat belt and if air bag deploy is -6750N

(b) The net force that stop the person if he/she is not restrained is -1.35×106N

See the step by step solution

Step by Step Solution

Step 1 : Given Information

Mass of the personm=60kg

Initial velocity of the carvi=15m/s

Step 2 : Calculation of A

Formula used :

The constant acceleration, the third equation of the kinematics is used. The initial and final velocities can be related with the following relation.

vf2 =vi2 +2axI

The final velocity vfis zero,

Therefore acceleration is given by a =-vi22axII

The force acting on the passenger is given by

F = ma =-mvi22xIII

Conclusion :

When the air bag is deployed, and the seat belt is hold the stopping distance will be

x=1

using the equation III

F =ma =-mvi22x

Force can be calculated by substituting the values of mass of the person, initial velocity of the car and the stopping distance.

F =-60kg(15m/s)221 =-6750N

Negative sign indicates that it is pushing the passenger back to the seat.

The net force on the person if he/she is wearing a seat belt and if air bag deploy is-6750N

Step 3 : Calculation of B

When there are no restraints, then the stopping distance will be equal to

x=5 mm orx=0.005m

Using the equation III

F =ma=-mvi22x

Force can be calculated by substituting the values of mass of the person, initial velocity of the car and the stopping distance.

F =-60kg(15m/s)220.005 =-1.35×106N

This force is much larger than the restrained case.

Conclusion :

The net force that stops the person if he/she is not restrained is -1.35×106N

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