Acceleration given velocity and distance

[swede5670]

Homework Statement

A car traveling at 90 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.86 m. What was the average acceleration in g's of the driver during the collision?

Homework Equations

The Attempt at a Solution

The acceleration in m/s/s is -363.37 but I tried dividing by 9.8 and I got -37.05 g's but evidently that's wrong. I'm not sure what I'm missing here, I converted the km/h into m/s/s then got the correct acceleration then devided by 9.81. Is there a rule that I didn't use? Is there a limit on negative gs?

 

[jumbogala]

You are right, the acceleration is -363.37 m/s/s.

I'm not a hundred percent sure, but I think -37.08 g's is actually correct. Because if one g is worth 9.8 m/s/s, and you had positive 37.08 of them, your acceleration would be in the positive direction. But it's not... so I see no problem with having a negative number there.

Is there another reason you think -37 g's is wrong?

 

[baba89]

The average acceleration during the collision can be calculated by using the equation:

a = (v^2 - u^2) / (2s)

Where:
a = acceleration
v = final velocity (in m/s)
u = initial velocity (in m/s)
s = distance (in m)

In this case, the initial velocity is 90 km/h, which is equivalent to 25 m/s. The final velocity is 0 m/s since the driver comes to rest. The distance traveled is 0.86 m.

Substituting these values into the equation, we get:

a = (0^2 - 25^2) / (2*0.86)
a = -363.37 m/s^2

To convert this into g's, we can divide by the acceleration due to gravity, which is 9.8 m/s^2:

a = -363.37 / 9.8
a = -37.05 g's

There is no limit on negative g's, as this value represents the acceleration in the downward direction. It is important to note that the negative sign indicates that the acceleration is in the opposite direction of the initial velocity.