Calculating Car Collision Distance at 30 "g's"

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SUMMARY

The discussion focuses on calculating the distance required for a car to come to a complete stop during a collision with a deceleration of 30 "g's" (294 m/s²) from an initial speed of 100 km/h (27.78 m/s). The relevant equation used is v² = u² + 2as, where v is the final velocity (0 m/s), u is the initial velocity (27.78 m/s), a is the deceleration (-294 m/s²), and s is the distance. By rearranging the equation, the distance can be calculated as approximately 0.5 meters, indicating the design requirement for crumple zones in vehicles.

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candycooke
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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 30 "g's" (1.00 g = 9.80 m/s^2). Assuming uniform deceleration at this rate, calculate the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 100km/h.

I'm not sure how to get this problem started, any help would be greatly appreciated.
 
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Acceleration is just change in speed divided by time.
You probably want the equation v^2 = u^2 + 2 a s
Where V is the final and u the intial speed, a is acceleration and s is distance.
Be careful about the sign of the acceleration.
 

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