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Q. 42

Expert-verifiedFound in: Page 156

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

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 surfaces, and the unfortunate occupant then decelerates over a distance of only about $5mm$. 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 $-1.35\times {10}^{6}N$

Mass of the person$\left(m\right)=60kg$

Initial velocity of the car$\left({v}_{i}\right)=15m/s$

**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.

${v}_{f}^{2}={v}_{i}^{2}+2a\u2206x\left(I\right)$

The final velocity $\left({v}_{f}\right)$is zero,

Therefore acceleration is given by $a=-\frac{{v}_{i}^{2}}{2a\u2206x}\left(II\right)$

The force acting on the passenger is given by

$F=ma=-\frac{m{v}_{i}^{2}}{2\u2206x}\left(III\right)$

**Conclusion :**

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

$\u2206x=1$

using the equation $\left(III\right)$

$F=ma=-\frac{m{v}_{i}^{2}}{2\u2206x}$

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

$F=-\frac{\left(60kg\right){(15m/s)}^{2}}{\left(2\right)\left(1\right)}=-6750N$

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$-6750N$

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

$\u2206x=5mmor\u2206x=0.005m$

Using the equation $\left(III\right)$

$F=ma=-\frac{m{v}_{i}^{2}}{2\u2206x}$

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

$F=-\frac{\left(60kg\right){(15m/s)}^{2}}{\left(2\right)\left(0.005\right)}=-1.35\times {10}^{6}N$

This force is much larger than the restrained case.

**Conclusion :**

The net force that stops the person if he/she is not restrained is $-1.35\times {10}^{6}N$

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