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

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SUMMARY

The average acceleration of a driver during a collision, where a car traveling at 85 km/h strikes a tree and comes to rest after traveling 0.80 m, is calculated to be -35.5 g's. The initial velocity was converted to 23.6 m/s, and using the formula (vf^2 - vi^2) / (2d), the acceleration was determined to be -348.39 m/s². This value was then expressed in terms of g's by dividing by 9.81 m/s², confirming the significant deceleration experienced during the collision.

PREREQUISITES
  • Understanding of kinematic equations, specifically (vf^2 - vi^2) / (2d)
  • Knowledge of unit conversion from km/h to m/s
  • Familiarity with the concept of acceleration in physics
  • Basic understanding of gravitational acceleration (1 g = 9.81 m/s²)
NEXT STEPS
  • Study the derivation and application of kinematic equations in collision scenarios
  • Learn about the effects of deceleration on the human body during car accidents
  • Research the physics of vehicle collisions and energy transfer
  • Explore advanced topics in dynamics, such as impulse and momentum
USEFUL FOR

Students studying physics, automotive safety engineers, and anyone interested in understanding the dynamics of vehicle collisions and their effects on occupants.

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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Looks right. Thats why accidents suck.
 

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