Kinematics?

Question

A car traveling at 93 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.90 m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's" where 1.00 g = 9.80 m/s^2

Answer 1

sorry, don't know....

Answer 2

from u=93kmph to v=zero in s=0.9m

equation of motion v^2=u^2 + 2fs

so deceleration f  =(93,000/3600)^2 / (2 * 0.9)

Divide that  by 9.8 to get no of gs

Answer 3

Vi=93km/h= 93000m/3600s=25.8m/s

Vf=Vi-a*t --> 0=Vi-a*t --> a*t=Vi

dx=vi*t-.5*a*t^2 --> dx=vi*t-.5*a*t*t --> dx=Vi*t-.5*Vi*t

solve for t

t=0.0349s

a=dv/dt

a=(0m/s-25.8m/s)/0.0349s =739m/s^2 or

a=739m/s^2 / 9.8m/s^2

so a=75g