A73-kg person in a moving car stops during a car collision in a distance of 0.80 m . the stopping force that the air bag exerts on the person is 8000 n. determine the speed of the person before the air bag opens up.

 v₀ = 13.24 m / s

Explanation:

Let's use Newton's second law to find the average acceleration during the crash

       F = m a

.       a = F / m

       a = 8000/73

       a = 109.59 m / s²

Now we can use the kinematic equations to find the initial velocity, since when the velocity stops it is zero (v = 0)

       v² = v₀² - 2 a x

       v₀² = 2 a x

       v₀ = √  2 a x

       v₀ = √ (2 109.59 0.80)

       v₀ = 13.24 m / s

answer: the answer is a

this would be about 43%