A car traveling 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. Part A What was the magnitude of the average acceleration of the driver during the collision?

Assuming the acceleration is constant, then we can use the derived equations for rectilinear motion. The equation is written below:

2ax = v² - v₀²

where

a is the acceleration

x is the distance

v is the final velocity

v₀ is the initial velocity

Since the car came to a stop, v = 0. Substituting the values,

2a (0.80 m) = 0² - [ (95 km/h) (1000 m/1 km) (1 h/3600 s) ]²

Solving for a,

a = - 435.23 m/s²

The sign is negative because it is decelerating.

Applying the formula

v² - v0² = 2as

a = (v² - v0²) / 2s

v = 0; v0 = 85 km/h = 85000/3600 m/s = 23.6 m/s; s = 0.80 m

a = 0 - (23.6) ² / 2*0.80 = - 348 m/s² = - 35.55g ≈ - 36g