Average G-Force Calculation for Abrupt Deceleration from 30mph in 1m Distance

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To calculate the average g-force experienced during abrupt deceleration from 30 mph to rest over 1 meter, the equations of motion are applied. The initial velocity (Vo) is 30 mph, which converts to approximately 13.41 m/s. Using the equation V^2 = Vo^2 + 2aΔS, where the final velocity (V) is 0 and ΔS is 1 m, the calculated acceleration (a) is -90.7 m/s². This acceleration corresponds to approximately -9.25 g's, indicating the average g-force experienced during the deceleration. The discussion highlights the importance of unit conversion and proper application of kinematic equations in solving the problem.
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Homework Statement


Determine the average number of g's experienced by someone abruptly coming to rest from an initial velocity of 30mph (such as in a car crash). Take the distance involved in coming to rest to be 1 m.

Homework Equations


V=Vo+at
ΔS=VoT+1/2at^2
V^2=Vo^2+2aΔS

The Attempt at a Solution


t=?
a=?
Vo= 30 mph
V=?
ΔS=?

Thank you, I'm really stuck and any help will be deeply appreciate it.
 
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You know Vfinal = 0 and ΔS (change in distance) = 1m. does that help?
 
so...
t=?
a=?
Vo= 30 mph
V=0
ΔS=1m

V^2=Vo^2+2aΔS
0=(30mph)^2+2a(1m)
which resulted in a= -450mph

I know g= 9.8m/s ... so I'm assuming I have to change the -450mph to m/s by diving -450/3600s = 0.125m/s

I really don't know what to do next or if I'm in the right track...
 
V^2=Vo^2+2aΔS
0=(30mph)^2+2a(1m)
which resulted in a= -450mph
a should be in units of m/s2

(convert mph to m/s)
 

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