A car traveling 85 km/hr strikes a tree. The front end of the car compresses and the driver comes to rest...?

Question

A car traveling 85 km/hr strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 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.

I have tried this problem numerous times and keep getting -34 g as an answer when -36 g is listed as the correct answer.

Thanks for all your help!

Answer 1

I did it sing the equation

v^2=v0^2+2ad

and I considered the need to transfer the spped unit so I can tranfer to grams

and I got 35.55 which is a pproximatlu 36

Answer 2

vo = 85 km/hr = 85000 m / 3600 sec

vot – ½at2 = 0.80 meters

vo – at = 0 m/s

Solving for t:

vo – at = 0 m/s

vo = at

t = vo / a

Substitute t into the first equation

vot – ½at2 = 0.80 meters

vo(vo / a) – ½a(vo / a)2 = 0.80 meters

vo2/ a – ½a(vo2 / a2)  = 0.80 meters

vo2/ a – ½(vo2 / a) = 0.80 meters

½(vo2 / a) = 0.80 meters

(vo2 / a) = 1.60 meters

(vo2 / 1.60 meters) = a

(85000 meters / 3600 sec) 2 / 1.60 meters = a

a = 348.43 m/s2 or 350.00 m/s2 (two significant figures)

(350 m/s2) / (9.80 m/s2) = 35.71 g

or 36 g (two significant figures)

Answer 3

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