What Was the Driver's Average Acceleration During a Collision?

  • Thread starter High_Voltage
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In summary, the driver of a car traveling at 85km/h collides with a tree, compressing the front end of the car and coming to rest after traveling 0.80m. Using the equation (vf^2-vi^2)/(2d)=a, the average acceleration of the driver during the collision is calculated to be -35.51g's in the negative direction. This highlights the dangers of accidents.
  • #1
High_Voltage
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Homework Statement


A car traveling 85km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's" where 1.00g = 9.81m/s^2.


Homework Equations



(85 km/h)(1000m/1km)(1h/3600s)= 23.6m/s

(vf^2-vi^2)/(2d)=a


The Attempt at a Solution



vi=23.61m/s
d=.80m
vf=0m/s

(0^2)-(23.6^2)/(2x.80)= -348.39m/s^2

-348.39/(9.81)= -35.51g's 35.5g's in the negative direction.

This does not seem right to me, can anyone confirm or lead me in the right direction, not do the work for me as I know that is frowned upon here =)
 
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  • #2
Looks right. Thats why accidents suck.
 

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