Question

A car traveling 93 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.90 m Part A What was the magnitude of the average acceleration of the driver during the collision? Express your answer using two significant figures a= m/s
Part B Express the answer in terms of "g's," where 1.00g 9.80 m/s2 Express your answer using two significant figures.

Answer 1

Answer:

A) $a=14.35m/s^2$

B) $a = 1.46g$

Explanation:

First we need to convert car's speed:

Vo = 93 km/h * 1000m / 1km * 1h / 3600s = 25.83m/s

Now we can calculate the acceleration with:

$Vf^2 = Vo^2 - 2*a*d$   Solving for a:

$a = \frac{Vf^2 - Vo^2}{2*d}$    where Vf=0m/s   Vo = 25.83m/s   d=0.9m

$a=14.35m/s^2$

If we divide this value by gravity = 9.8, we get:

$a=1.46g$