# Severity Index and Collision Distance Calculation

## What is the severity index for the collision?

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## How far does the person travel during the collision if the car was initially moving forward at 4.30 km/h?

Final answer:

The person travels a distance of approximately 0.1958 meters during the **collision**. The **severity index** cannot be calculated without the time (t) of the acceleration.

To calculate the severity index (SI) for the collision, we need to determine the *acceleration* in multiples of g and the time the acceleration lasts.

Given:

- Acceleration (a) = 35.0 m/s^2
- Initial
*velocity*(u) = 4.30 km/h - Final velocity (v) = 14.0 km/h

First, let's convert the initial and final velocities to m/s:

- Initial velocity (u) = 4.30 km/h = 1.19 m/s
- Final velocity (v) = 14.0 km/h = 3.89 m/s

Next, we can calculate the acceleration:

Using the equation v^2 = u^2 + 2as, we can solve for acceleration (a):

3.89^2 = 1.19^2 + 2a * s

15.12 = 1.4161 + 2a * s

2a * s = 13.7039

a * s = 6.85195

Since we know the acceleration (a) is 35.0 m/s^2, we can solve for s:

35.0 * s = 6.85195

s = 0.1958 m

Therefore, the person travels a distance of approximately 0.1958 meters during the collision.

To calculate the severity index (SI), we can use the formula SI = a^(5/2) * t:

SI = (35.0)^(5/2) * t

Since the time (t) is not given in the question, we cannot calculate the severity index without this information.