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forsale [732]
3 years ago
9

A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the dec

eleration 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?
Physics
1 answer:
postnew [5]3 years ago
8 0

Answer:

Distance = 0.738 m

Explanation:

Solution:  

<u> First convert Km/h into m/s.</u>  

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

<u>According to third equation of motion: </u>

Vf^{2} – Vi^{2} = 2 * acceleration * distance

<em> Vf= final velocity </em>

<em>Vi= initial velocity</em>

putting values in third equation of motion....

0.2^{2} – 20.8333^{2} = 2 * (-30 * 9.8) * distance

note:

negative sign is due to deceleration

Distance = 0.738 m

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