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During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of 60 g, but lasting for only 36 {\rm ms} (or less).

How far (in meters) does a person travel in coming to a complete stop in 36 {\rm ms} at a constant acceleration of 60g?

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Answer:

0.3814128 m

Explanation:

t = Time taken = 36 ms

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 60g

g = Acceleration due to gravity = 9.81 m/s²

[tex]v=u+at\\\Rightarrow u=v-at\\\Rightarrow u=0-(-9.81\times 60)\times 36\times 10^{-3}\\\Rightarrow u=21.1896\ m/s[/tex]

[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-21.1896^2}{2\times -9.81\times 60}\\\Rightarrow s=0.3814128\ m[/tex]

The distance the person traveled is 0.3814128 m

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