Question

Very 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 ( ), defined as =5/2 . In the expression, is the duration of the accleration, but is not equal to the acceleration. Rather, is a dimensionless constant that equals the number of multiples of that the acceleration is equal to.

In one set of studies of rear-end collisions, a person's velocity increases by 12.1 km/h with an acceleration of 35.0 m/s2 . Let the + direction point in the direction the car is traveling. What is the severity index for the collision?

How far does the person travel during the collision if the car was initially moving forward at 5.80 km/h ?

Answer 1

SI = 2.31

d = 0.161 m

Step-by-step explanation:

First, convert km/h to m/s.

12.1 km/h × (1000 m/km) × (1 h / 3600 s) = 3.36 m/s

Find the time.

v = at + v₀

3.36 = (35.0)t + 0

t = 0.0960

Find the value of a.

a = a / g

a = 35.0 / 9.8

a = 3.57

Solve for SI.

SI = (3.57)^(⁵/₂) (0.0960)

SI = 2.31

Solve for d.

v² = v₀² + 2aΔx

(3.36)² = 0² + 2 (35.0) d

d = 0.161