Question

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of 60 g, but lasting for only 31ms (or less).

How far (in meters) does a person travel in coming to a complete stop in 31ms at a constant acceleration of 60g?

Answer 1

Let's see what we have here:
\[a=60g = (60)(-9.8m/s^2)=-588m/s^2\]\[\Delta t = 31ms=0.031s\]\[V_f = 0m/s\]\[V_i=?\space m/s\]\[d=? \space m\]
Now use the kinematic equation:\[V_f=V_i+at\]\[0=V_i+(-588m/s^2)(.031s)\]\[V_i=18.228m/s\]Now that you know the initial velocity, you can calculate the distance:\[d=V_0t+\frac{1}{2}at^2=(18.228m/s^2)(0.031s)+\frac{1}{2}(-588m/s^2)(0.031s)^2\]You should end up with:\[d=0.28m\]

Answer 2

Note that there are a couple of ways to do this one...this is just the first one I thought of.

Answer 3

Thanks, this is the way we are learning in my class I just get confused on where to plug in numbers and how to solve for certain ones. I came up with .028m before so I must have been off by just a little. I really appreciate the help!

Answer 4

np :)

Answer 5

I have one more homework problem I am working on, since yesterday night and I can't seem to solve it, I think I am close but maybe you could help me on this last one?

If a pilot accelerates at more than 4 g, he begins to "gray out," but not completely lose consciousness. What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.6 (3.6 times the speed of sound) without graying out? (Use 331m/s for the speed of sound in cold air.)

I end up with 34.6880 s, but it keeps telling me I'm wrong and I'm not sure how else to figure it out!

Answer 6

It seems like I answered a very similar question a few days ago...lemme look

Answer 7

I just looked over my notes again and I think maybe I found my error.
I think to solve for time it would look like:
(1191.6 - 0)/(35.316) = 33.74 s

Answer 8

It's very similar...maybe this will help:

http://openstudy.com/users/shane_b#/updates/522fb485e4b03eb771a1d07f

Your answer may be right (I haven't checked it)...so you may want to make sure you adjusted for significant figures first.

Answer 9

Okay, thanks :)

Answer 10

Did you find your mistake? I can work it out if you like...

Answer 11

Well I looked at my answers before and I guess I already tried 34.7 s, so I'm looking at that other problem and trying to solve again. Would I start by using
t = V(final) - V(initial)/a ?

Answer 12

That should work :)

Answer 13

Well for acceleration I'm using (3.6 x 9.8 m/s^s). I think that's right? And then for my velocity final would I use 4? And initial 0? I'm going wrong somewhere.

Answer 14

30.4s is correct, but the numbers I'm plugging in aren't correct.

Answer 15

I was using 4g as a max acceleration

Answer 16

Ok, maybe I read it right the first time :) Let me read it again slowly...one sec.

Answer 17

Yea, had it right the first time:

\[\frac{1191.6m/s - 0m/s}{(39.2m/s^2} = 30.4s\]

Answer 18

How did you find the 39.2?

Answer 19

Your acceleration should be 4*9.8m/s^2 (which = 4g)

Answer 20

That's the max acceleration the pilot can withstand.... 4gs

Answer 21

Oh, okay, I was doing 3.6 x 9.8

Answer 22

Yep, I see that. You already used 3.6 to get the final velocity anyway so it wouldn't make sense to use it again!

Answer 23

Thanks very much for the help, I just get so confused with all the numbers! My professor doesn't explain very well, just gives notes on a PowerPoint so I'm trying to figure out how to solve these problems, without ever having taken a physics course before. I appreciate the assistance :)

Answer 24

It's no problem at all. I even confuse myself sometimes and I'm generally pretty good with the kinematic stuff :)

Physics, just like math, is all about practice, practice, practice!

Answer 25

You could also check out the physics videos at khanacademy.org

Very helpful :)

Answer 26

Yeah I will definitely be practicing many problems before my exam! I will have to check that out, thanks for the info! :) Hopefully with all this help/input I can start solving more of the problems on my own. Thanks again!

Answer 27

np :)