HOW do you solve these problems? (Physics)

[The4Aces]

The human body can survive a negative acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 243 m/s2. If you are in an automobile accident with an initial speed of 127.0 km/h and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

Earthquakes produce several types of shock waves. The most well-known are the P-waves and the S-waves. In the earth’s crust, the P-waves travel around 7.0 km/s while the S-waves move at about 3.6 km/s. The time delay between the arrival of these two waves at a seismic recording station tells geologists how far away the earthquake occurred. If the time delay is 19 s, how far away from the seismic station did the earthquake occur?


I have spennt the last 2 hours trying to figure this out and i cant.

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The human body can survive a negative acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 243 m/s2. If you are in an automobile accident with an initial speed of 127.0 km/h and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

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Fall semester, eh? http://forumserver.twoplustwo.com/showflat.php?Cat=&Number=3333445&page=1&view=colla psed&sb=5&o=&fpart=1


[ QUOTE ]
Earthquakes produce several types of shock waves. The most well-known are the P-waves and the S-waves. In the earth’s crust, the P-waves travel around 7.0 km/s while the S-waves move at about 3.6 km/s. The time delay between the arrival of these two waves at a seismic recording station tells geologists how far away the earthquake occurred. If the time delay is 19 s, how far away from the seismic station did the earthquake occur?

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If the time delay was 1sec, then what was it?

You didn't really spend two hours on this.

[The4Aces]

i feel like a dumbass but i seriously cant figure out the one second thing.

I don't want to do your homework for you, but here are some equations from high school physics:

Question 1:

v² = u² + 2as

where v = final velocity, u = initial velocity, s = distance.

Question 2:

Very easy if you think about it. Every second, the P-wave travels (7.0 - 3.6) = 3.4km more. So, in 19 seconds...

Is this a joke post?

[The4Aces]

I know the awnser for the 2nd one is 140.8 because its in the back of the book, but i seriously can not figure that [censored] out.

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i feel like a dumbass but i seriously cant figure out the one second thing.

[/ QUOTE ]You're good. I'm convinced you're not joking. John walks 3 ft/s. Tom walks 2 ft/s. They start at the same point. If the delay between John & Tom reach you is 1 sec, what is the distance they started at? Now do 19 seconds. Now solve the problem.

[The4Aces]

I seriously dont get the last problem. I know you probably dont belive me becuase it is coming to easy to you. But seriously i can not get 140.8 no mater what it try.

OK, fine.

Assume the distance to the epicenter is D. At time T, the P wave arrives, having traversed (T*7)km. At time T+19, the S wave arrives, having traversed (T+19)*(3.6)km

These two distances are equal to D. So we have:

(T+19)*3.6 = T*7

I'll leave the rest up to you.

[The4Aces]

your the man. that makes me feel really stupid that i spent so long trying to figure that out.

Nah it's a tough subject. I screwed it up too when I first looked at it, and I have a four year degree in it.

There is a more intuitive way of looking at it:

At the time when the P wave reaches the station, it has gained 19s * 3.6 km/s = 68.4km on the S wave. The P wave moves at 7km/s and opens up a gap of 3.4 km/s. So the ratio of distance moved/gap opened up is 7/3.4 = 2.058km per km in gap. Therefore the total distance it has travelled is 64.8*2.058 = 140.8km.