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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 31ms (or less).


How far (in meters) does a person travel in coming to a complete stop in 31ms at a constant acceleration of 60g?

Verified Answer :

d=0.28mVғ = final speed of passenger = 0 
d = stopping distance = ? 
a = acceleration = - 60 • (9.8) = - 588 m/sec² ... negative value for deceleration 

31ms = 0.031 sec 

  Vғ = (a • t) + Vi 
   0 = (- 588 • 0.031) + Vi 
    Vi = 18.228m/sec 

              (Vғ)² – (Vi)² = 2 • a • d 
            (0)² – (18.228)² = 2 • (- 588) • d 

                  d = 0.028meters

an other way to solve
Let's see what we have here:
a=60g=(60)(−9.8m/s2)=−588m/s2Δt=31ms=0.031sVf=0m/sVi=? m/sd=? m
Now use the kinematic equation:Vf=Vi+at0=Vi+(−588m/s2)(.031s)Vi=18.228m/sNow that you know the initial velocity, you can calculate the distance:d=V0t+12at2=(18.228m/s2)(0.031s)+12(−588m/s2)(0.031s)2
d=0.28m

Answer: 0.28 m

Deceleration, a= -60 g

where g is the acceleration due to gravity\(=9.8 m/s^2\)

So, \(a= -60\times 9.8 m/s^2=-588 m/s^2\)

Using the equation of motion, we need to find the initial velocity,

\(v-u=at\)

final velocity, \(v=0\)

time, \(t= 31 ms=31\times 10^{-3}s\)

\(0-u=-588m/s^2\times 31\times 10^{-3} s=18.23 m/s\)

now, using third equation of motion,

\(s=ut+\frac{1}{2}at^2\)

we can find out the distance traveled by the person before coming to complete stop.

\(s=18.23m/s\times 31\times 10^{-3}s+\frac{1}{2}(-588 m/s^2) \times (31\times 10^{-3} s)^2=0.56 m-0.28 m=0.28 m\)


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