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A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed about 30 "Gs" (1.0 g = 9.8 m/s^2). Assuming uniform deceleration of this value, what is the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 75 km/h?

Respuesta :

Hashirriaz830

Answer:

Distance = 0.738 m

Explanation:

Solution:  

First convert Km/h into m/s.  

75 km/h * 1000 m/km * 1 hr/3600 sec = 20.8333 m/s  

According to third equation of motion:

[tex]Vf^{2}[/tex] – [tex]Vi^{2}[/tex] = 2 * acceleration * distance

Vf= final velocity

Vi= initial velocity

putting values in third equation of motion....

[tex]0.2^{2}[/tex] – [tex]20.8333^{2}[/tex] = 2 * (-30 * 9.8) * distance

note:

negative sign is due to deceleration

Distance = 0.738 m

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