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 m. what is the force exerted on a 65.0-kg driver who decelerates from 18.0 m/s to rest over a distance of 1.00 m?

Question

Juliana

Physics

27 March, 21:42

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 m. what is the force exerted on a 65.0-kg driver who decelerates from 18.0 m/s to rest over a distance of 1.00 m?

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Answers (1)

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Kiara Clarke · Answer 1

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

Thus,

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