the combination of crumple zones and air bags and seat belts might increase the distance over which a person stops in a collision to as much as 1.00 mm . what is the force exerted on a 65.0- kgkg driver who decelerates from 18.0 m/sm/s to rest over a distance of 1.00 mm ?



Answer :

From Newton's second law of motion,

F = ma

We know m = 65 kg. Next, we have to find a through this formula:

v² = v₀² + 2ad

0² = (18 m/s)² + 2(a)(1 m)

Solving for a,

a = -162 m/s²

So,

F = (65 kg)(-162 m/s²) = -10,530 Newtons.

A crumpled zone in the front of the car gradually slows the car down and increases the braking time. A safety cell is a rigid cage that prevents passengers from becoming trapped. Inside the car seat belts and airbags prevent the driver from hitting the windshield steering wheel and dashboard.

Crumple zones are typically found in the front and rear of a vehicle and are designed to crumple and crush under high-force impacts. The Crumple Zone transfers a portion of the vehicle's kinetic energy during impact to a controlled crash while maintaining the integrity of the occupant cell.

Learn more about Crumple zones here:-https://brainly.com/question/26544925

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