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?

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

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.