Answer:
a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.
Explanation:
a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:
[tex]a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]a' = 3.569[/tex]
The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:
[tex]v_{f} = v_{o} + a \cdot t[/tex]
Where:
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v_{f}[/tex] - Final speed, measured in meter per second.
[tex]a[/tex] - Acceleration, measured in [tex]\frac{m}{s^{2}}[/tex].
[tex]t[/tex] - Time, measured in seconds.
[tex]t = \frac{v_{f}-v_{o}}{a}[/tex]
[tex]t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }[/tex]
[tex]t = 0.095\,s[/tex]
Lastly, the severity index is now determined:
[tex]SI = \frac{a'^{5}}{2\cdot t}[/tex]
[tex]SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}[/tex]
[tex]SI = 3047.749[/tex]
b) The initial and final speed of the injured are [tex]1.944\,\frac{m}{s}[/tex] and [tex]5.278\,\frac{m}{s}[/tex], respectively. The travelled distance can be determined from this equation of motion:
[tex]v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s[/tex]
Where [tex]\Delta s[/tex] is the travelled distance, measured in meters.
[tex]\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s = 0.345\,m[/tex].
