Question

ery large accelerations can injure the body, especially if they last for a considerable length of time. One model used to gauge the likelihood of injury is the severity index ( S????SI ), defined as S????=a5/2tSI=a5/2t . In the expression, tt is the duration of the accleration, but aa is not equal to the acceleration. Rather, aa is a dimensionless constant that equals the number of multiples of gg that the acceleration is equal to. In one set of studies of rear-end collisions, a person's velocity increases by 16.4 km/h16.4 km/h with an acceleration of 34.0 m/s234.0 m/s2 . Let the +x+x direction point in the direction the car is traveling. What is the severity index for the collision?

Answer 1

Answer:

2.98 second

Explanation:

The severity index is defined by :

$S=a^{5/2}t$

a is dimensionless constant that equals the number of multiples of g

Conditions are given as :

Initial velocity, u = 0

Acceleration, a = 34 m/s²

Final velocity, v = 16.4 km/h = 4.56 m/s

We can find t from the above data as follows :

$t=\dfrac{v-u}{a}\\\\t=\dfrac{4.56-0}{34}\\\\t=0.134\ s$

As a is the acceleration that is multiple of g.

So,

$a=\dfrac{34}{9.8}=3.46$

So,

Severity index,

$S=a^{5/2}t\\\\S=(3.46)^{5/2}\times 0.134\\\\S=2.98\ s$

Hence, the severity index for the collision is 2.98 seconds.