What is the g-force of a car stopping on a dime?

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SUMMARY

The g-force experienced by a car traveling at 55 mph when stopping across the diameter of a dime is calculated to be approximately 1721g. This value is derived using the formula for acceleration, a = v^2 / (2d), where v is the initial velocity converted to meters per second and d is the stopping distance of 17.91 mm. The calculations assume constant acceleration throughout the stopping process. The methodology involves converting speed from mph to mps and applying kinematic equations to determine the resulting g-force.

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What is the g-force on a car, initially traveling at 55 mph, when stopping across the diameter of a dime?
 
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convert mph to mps, find the acceleration from change in velocity, multiply by mass, divide by g.
 
Find acceleration from change in velocity over given distance, that is.
 
Would it not be appropriate to solve a system of equations, d=1/2 * a * t^2, d = 17.91 mm, and a * t = 55 mph?

Doing this I get about 16.9 kilometers / s^2, or 1721g.
 
Last edited:
a = \frac{v^2 - v_0^2}{2d}

but v_0 = 0 (in my ref frame : )

so

a = \frac{v^2}{2d}

Of course, this assume constant acceleration...
 

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