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 (SI), defined as SI=a5/2t . In the expression, t is the duration of the accleration, but a is not equal to the acceleration. Rather, a is a dimensionless constant that equals the number of multiples of g that the acceleration is equal to. In one set of studies of rear-end collisions, a person's velocity increases by 16.9 km/h with an acceleration of 34.5 m/s2 . Let the +x direction point in the direction the car is traveling. What is the severity index for the collision?

Answer 1

Severity index= a(superscript 5/2)t
to solve for a: 34.5 m/s2 ---> since g = 9.8 m/s^2, a = (34.5/9.8) = 3.52
= (3.52)^(2.5) times (t)
= 23.25t

now, to solve for time:
under constant acceleration: velocity final= velocity initial + (acceleration x time)
change in velocity= 16.9 km/hour = (velocity final- velocity initial)
16.9 km/hour= 16900 meters/3600 seconds= 4.69 meters/second= (velocity final - velocity initial)
thus, (velocity final-velocity initial) = (acceleration x time)
4.69= (34.5m/s2)(time)
4.69= 34.5t
time = 0.14 seconds

thus, severity index = 23.25t=23.25 (0.14)= 3.255

(This is solved with approximation at each step, so look into the following attachment in which approximation is done in last step only):

Let me know if you have any questions or need clarification.