G's of acceleration?

Question

I think that g's of acceleration are just a/9.8, right? Not sure how to set up this problem...

You read about a car accident in the newspaper. A car traveling at 70 km/h collided with a concrete bridge support. The front end of the car was compressed 0.92 m, and the car came to a complete stop within this distance.

How many g's of acceleration did the occupant of the car experience?

I feel like I need to know his mass or something, but no other info is given??

thank you!!! I forgot to square the "19.44"!!!!

thank you!!! I forgot to square the "19.44"!!!!

Answer 1

First, you have to assume that the decelleration was constant. Next, you know the car traveled .92 m during decelleration. It started at 70km/h (=70,000meters/h=19.44m/s) and ended at 0, using this motion equation:
vf^2=vo^2+2a(delta x)
you can now solve the problem, with vf=0, vo=19.44, and dx=.92.
Plugging in you get
0=377.91+2a*.92
-377.91=1.84a
-205.39=a
Divide by 9.8 to get g's:
**-20.96**
(That's one crazy crash!).

Answer 2

G's mean the number of g = 9.81 m/sec^2 in an acceleration a there are. Thus, G = a/g is correct. So you need to calculate the a, acceleration, as the car comes to a screeching halt against that bridge.

So here's what we know. The car went from u = 70 kph to v = 0 kph in d = .92 m. Drawing on the SUVAT equation v^2 = u^2 + 2ad, we solve for a = (v^2 - u^2)/2d = -(u^2)/2d. Change u = 70 kph into meters per second and you can solve for a. Then G = a/g will give you a in number of G's.

And, lo, no mass is needed. But, and this is a big BUT, you would need mass m of the driver if you wanted to know how much force the hapless driver was subjected to. That results because f = ma, which is the force on the driver decelerating at a or G = a/g.

Answer 3

you would only need the mass if you were looking at the forces involved.
since they are asking for acceleration you are only concerned with the change in speed.
the initial speed was 70km/hr. you'll convert that to m/s.
the distance it stopped in is given as .92m.
remember that the distance traveled during constant acceleration (or deceleration in this case) is given by
Vi=at and d=Vave t, so by substitution
Vi = a x d / Vave. (where V average = Vinitial / 2)
Rearrange and
a = Vi^2 / 2d
that will give you the acceleration, and the number of g's is the acceleration / 9.8

Answer 4

x= 0 m
xo= .92 m
Vo= 70 m/s
V= 0 m/s
a= ?
t=?

solve for time. then solve for acceleration.