Math Problem!!!! - Page 2 - GoFuckYourself.com

06-30-2003, 10:07 PM

Quote:

Originally posted by Mr.Wong

Didn't even see this. This is also correct. The train will travel -900m. How long would the train travel total? 1200 miles.

What's 1200-900=300 meters.

bdjuf where do you even address the time 30 seconds in your equation?

the graph shows that it would stop after 20 seconds, because of the deceleration speed

the extra 10 seconds would be 10 seconds of rest, in this case

if there WAS no context, then yes, it woulda been 300m because he woulda went 100m backwards

but since there IS context, its 400m

Mr.Wong · 06-30-2003, 10:08 PM

Quote:

Originally posted by smack
i'm coming up with 900 meters and here's why, instead of using vector analysis i used D= 1/2*(A*T^2)
D is the distance you're looking for, a is the absolute value of your acceleration and T is the time (30 seconds)

it's the same formula you use to calculate how fast something will fall with gravity except you replace the A with the gravitational constant.

that makes any sense or am i just high?

Yes, that's right. BUT it reads 0 + 1200 - 900 = 300meters. Luc, just fucking write down 300 meters. It's fucking right.

06-30-2003, 10:09 PM

Quote:

Originally posted by Mr.Wong

Yes, that's right. BUT it reads 0 + 1200 - 900 = 300meters. Luc, just fucking write down 300 meters. It's fucking right.

wanna bet?

phogirl69 · 06-30-2003, 10:10 PM

Quote:

Originally posted by Lane
omg, i knew this stuff at 9th or 10th grade or something..

how many of you are high school dropouts?

I don't ever recall doing these kinds of problems in high school.
What kind of math is this, Calculus???

06-30-2003, 10:11 PM

Quote:

Originally posted by phogirl69

I don't ever recall doing these kinds of problems in high school.
What kind of math is this, Calculus???

physics class in highschool and college

Mr.Wong · 06-30-2003, 10:16 PM

Quote:

Originally posted by bdjuf

wanna bet?

Prove it with an equation. You have mine and it checks. You have 2 others that also check. Your graph doesn't even match up to what you're saying. You have X as the unknown which you are using as time, but your also using the slope as time? You can't use both sides as time!

06-30-2003, 10:19 PM

Quote:

Originally posted by smack
i'm coming up with 900 meters and here's why, instead of using vector analysis i used D= 1/2*(A*T^2)
D is the distance you're looking for, a is the absolute value of your acceleration and T is the time (30 seconds)

it's the same formula you use to calculate how fast something will fall with gravity except you replace the A with the gravitational constant.

that makes any sense or am i just high?

that is a formula for the answer in Newton/second, you cant apply that in this context, unless you want to find the force of the train or something

06-30-2003, 10:21 PM

Quote:

Originally posted by Mr.Wong

Prove it with an equation. You have mine and it checks. You have 2 others that also check. Your graph doesn't even match up to what you're saying. You have X as the unknown which you are using as time, but your also using the slope as time? You can't use both sides as time!

there is no equation because it is not a theoretical question! its a PRACTICAL question, of a specific situation!
that is why we use graphs!

please tell me the flaws on my graphs for me to believe you

Luc Duboi · 06-30-2003, 10:22 PM

Quote:

Originally posted by Mr.Wong

Yes, that's right. BUT it reads 0 + 1200 - 900 = 300meters. Luc, just fucking write down 300 meters. It's fucking right.

BDjuf has grahp supports ....... I don't think he's wrong.

still waiting!!!

06-30-2003, 10:24 PM

Quote:

Originally posted by Luc Duboi
BDjuf has grahp supports ....... I don't think he's wrong.

still waiting!!!

thats exactly it, in their equation, it includes the train going BACKWARDS!!!!

want proof? replace the 30 seconds with 500 seconds, they would come up with a negative number

BlueDesignStudios · 06-30-2003, 10:24 PM

Quote:

Originally posted by bdjuf
here, I made a pic for you

So basically to summarize it all you're saying you disagree with all of Einstein's theories and in particular you dispute his papers on relativity?

06-30-2003, 10:24 PM

Quote:

Originally posted by BlueDesignStudios

So basically to summarize it all you're saying you disagree with all of Einstein's theories and in particular you dispute his papers on relativity?

this has nothing to do with einstein and relativity! its the way you proceed to solve these problems!!!

BlueDesignStudios · 06-30-2003, 10:33 PM

Quote:

Originally posted by bdjuf

this has nothing to do with einstein and relativity! its the way you proceed to solve these problems!!!

Technically it may have something to do with einstein's theory of relativity - it all depends on the location & velocity of the time keeper relative to the direction of the train's travel.
I guess to have the question layed out better, the time keeper should be positioned in the train to ensure we can rule out einstein's relativity theory interfering with the answer

06-30-2003, 10:48 PM

alright, Im hoping to make this my last post on this thread with a more comprehensible (I hope) graph that I just made

I hope this helps you understand better

fr33s3x · 06-30-2003, 11:01 PM

by the graph we see that
after 20 sec, the train will stop, and it doesnt matter if you add an extra 10, the distance wont change.

v0=40m/s
a=-2m/s^2
t=20s
----------
v=?
x=?
----------
v=v0-at (- because the acceleration is negetive)
v=40-2*20
v=40-40
v=0 - the speed after 20 sec
--------------------------------------
x=x0+v0t-at^2/2
x=40*20-2*20^2/2
x=800-400
x=400
-------------------------------------
the train will travelled 400meter within 20 secs, thats why the train travelled the same distance with 30 sec..
hope i helped.

06-30-2003, 11:02 PM

Quote:

Originally posted by fr33s3x
by the graph we see that
after 20 sec, the train will stop, and it doesnt matter if you add an extra 10, the distance wont change.

v0=40m/s
a=-2m/s^2
t=20s
----------
v=?
x=?
----------
v=v0-at (- because the acceleration is negetive)
v=40-2*20
v=40-40
v=0 - the speed after 20 sec
--------------------------------------
x=x0+v0t-at^2/2
x=40*20-2*20^2/2
x=800-400
x=400
-------------------------------------
the train will travelled 400meter within 20 secs, thats why the train travelled the same distance with 30 sec..
hope i helped.

well done! and thanx for proving my point

High Quality · 06-30-2003, 11:02 PM

hey dumb ass, 2m/s/s is an acceleration not a speed.

06-30-2003, 11:03 PM

Quote:

Originally posted by High Quality
hey dumb ass, 2m/s/s is an acceleration not a speed.

yeah? and Ive been treating it as a acceleration this whole time?
whats your point?

fr33s3x · 06-30-2003, 11:14 PM

guys!
bdjuf is fucking right!
the train will stop after 20 SEC, here is the proof:

v=v0-at (- because the acceleration is negetive)
ok!
we need to know when the speed will be 0, so we will see how much time it takes!

so:
v0=40m/s
a=-2m/s^2
v=0
-----
t=?
----
0=v0-at
0=40-2t
-40=-2t
t=20sec
the train will stop after 20 sec, thats it! no more argue!

Planet Bob · 06-30-2003, 11:14 PM

380 - thank you for playing

eroswebmaster · 06-30-2003, 11:15 PM

What's the deal, your dungeons and dragons groups not show up tonight?

phogirl69 · 06-30-2003, 11:15 PM

Quote:

Originally posted by bdjuf

physics class in highschool and college

I don't think they had physics at my high school, I think it was must have been an Honors class or something like that. I remember Chemistry and Biology though...

phogirl69 · 06-30-2003, 11:17 PM

Quote:

Originally posted by phogirl69

I don't think they had physics at my high school, I think it was must have been an Honors class or something like that. I remember Chemistry and Biology though...

I HATE math/chemistry equations crap. This shit gives me a fucken headache just looking at the equations

eroswebmaster · 06-30-2003, 11:17 PM

Quote:

Originally posted by phogirl69

I don't think they had physics at my high school, I think it was must have been an Honors class or something like that. I remember Chemistry and Biology though...

they had general physics in my highschool in the 9th grade.

but we were all too busy putting spit wads up our noses and flicking them onto Theresa mantzke's lips.

nosey · 06-30-2003, 11:17 PM

BlueDesignStudios · 06-30-2003, 11:21 PM

Could we have someone write a php script to demonstrate? People here seem to love writing php scripts

06-30-2003, 11:23 PM

Quote:

Originally posted by BlueDesignStudios
Could we have someone write a php script to demonstrate? People here seem to love writing php scripts

I dont think there is anything left to demonstrate

I gave the graphical approach
and freesex gave the theoretical approach

this thread should be closed

FINAL ANSWER: 400 meters

BlueDesignStudios · 06-30-2003, 11:24 PM

Quote:

Originally posted by bdjuf

I dont think there is anything left to demonstrate
I gave the graphical approach
and freesex gave the theoretical approach

this thread should be closed

FINAL ANSWER: 400 meters

Ok then just for completion.. how about you go over any assumptions made in coming up with your answer?

High Quality · 06-30-2003, 11:25 PM

Quote:

Originally posted by High Quality
hey dumb ass, 2m/s/s is an acceleration not a speed.

I wasn't talking to you. I was referring to the original thread. I have degrees in engineering and physics and this is a high school problem

Your analysis, while the long and drawn out, is the correct answer.

06-30-2003, 11:26 PM

Quote:

Originally posted by High Quality

I wasn't talking to you. I was referring to the original thread. I have degrees in engineering and physics and this is a high school problem

Your analysis, while the long and drawn out, is the correct answer.

good to hear that, and thanx for the feedback

06-30-2003, 11:28 PM

Quote:

Originally posted by BlueDesignStudios

Ok then just for completion.. how about you go over any assumptions made in coming up with your answer?

hehe, observe the final graph, and fr33s3x's final post

there is ALL you need, the graph alone is sufficient, but some people NEED to see the equation..

I just want mr.wrong to know, this was NOT a PLUG AND PLAY equation problem, there was 1 step he forgot to do, wich is find out that the train had stopped at 20 seconds

phogirl69 · 06-30-2003, 11:41 PM

Are you you guys sure this is a high school problem?

Don't ever recall doing anything like that and it looks difficult to me even now, lol.

Perhaps I did take Physics in high school, I really don't remember since it was over 10 years ago and I always HATED math/chemistry/equations.

I was going to go back to college but screw that if I have to do equations and graphs like that!! lol

06-30-2003, 11:59 PM

Quote:

Originally posted by phogirl69
Are you you guys sure this is a high school problem?

Don't ever recall doing anything like that and it looks difficult to me even now, lol.

Perhaps I did take Physics in high school, I really don't remember since it was over 10 years ago and I always HATED math/chemistry/equations.

I was going to go back to college but screw that if I have to do equations and graphs like that!! lol

haha! to be honest that was as simple as it got

High Quality · 07-01-2003, 12:02 AM

Ok here is an approach everyone can follow.

Speed (velocity) is miles/hour or distance / time.

--> Speed = distance / time
--> Distance = speed * time
--> Time = speed * distance

Initial speed is 40 meters / second.

Final speed is 0 meters / second

Acceleration is defined as a rate or in distance per second per second.

--> Acceleration = meters / second / second (m/(s^s))

Thus the equation is:

Speed initial - acceleration*(time) = Speed final
(m/s) - (m/s/s)*(s) = m/s <-- units check out
40m/s - 2m/s/s*(s) = 0m/s (note that s is unknown)
Solving we get S = 20seconds (deceleration time)

It will then stop within 20seconds, the 30s question is irrelevant.

Thus, to check the final distance:

Distance = speed * time OR
Distance = Initial speed*time - (acceleration*time*time)/2 (use calculus to prove this...)
(meters) = (meters/second)*seconds - (meters/second/second)*seconds*seconds)

--> Distance = (40m/s)*20seconds - (2m/s/s*20*20)/2

= 800meters - 400meters = 400meters distance.

07-01-2003, 12:05 AM

thank you for reconfirming my procedure

I await mr.wrongs reply on this thread

sexeducation · 07-01-2003, 09:16 AM

It depends on your definition of "brakes". To my understanding trains have two different types of "brakes". Friction and applied force by reversing the engines.

Therefore, there are two answers depending on the context of the question ....

Given ...
Acceleration =(FINAL VELOCITY - ORGINAL VELOCITY) divided by TIME

We get ...

-2MS^2 = (ZERO - 40/MS^2) / TIME

20 seconds of time before the train reaches zero velocity from the relative point in space for the question.

Therefore the train went the same distance it would accelerate forward at for 20 seconds ...
2/ms^2 for 20 seconds = 40 meters.

However, if the reverse engines are still being applied, which the question appears to be saying ... then
you need to add another 10 Seconds of acceleration and
we add another

20 Meters.

One answer is ...
The train traveled 60 Meters from the relative point in time in which the question was posed.

Quite simply 30 seconds of acceleration at 2/ms^2.

Dad@

sexeducation · 07-01-2003, 09:32 AM

Nope I screwed up the math ... lol

The answer is 20 seconds or 30 seconds of acceleration.

sexeducation · 07-01-2003, 09:36 AM

See what happens when you do not pay attention in Calculus class ...grrrr

lol

These questions are fun.

Porndealer · 07-01-2003, 10:01 AM

Battle of the minds

So what's the next question?

Mr.Wong · 07-01-2003, 11:29 AM

Quote:

Originally posted by bdjuf
thank you for reconfirming my procedure

I await mr.wrongs reply on this thread

bdjuf, I give it to you, well done. You're right, I wasn't taken into affect that the train was stopping. Damn, I used to be good at this in high school too.

Good job!

07-01-2003, 11:29 AM

Quote:

Originally posted by Mr.Wong

bdjuf, I give it to you, well done. You're right, I wasn't taken into affect that the train was stopping. Damn, I used to be good at this in high school too.

Good job!

thanx dude, it was fun debating you

fiveyes · 07-01-2003, 12:12 PM

Quote:

Originally posted by BlueDesignStudios

So basically to summarize it all you're saying you disagree with all of Einstein's theories and in particular you dispute his papers on relativity?

Otherwise known as the painless ice cream brain freeze, a common trap for the unwary at Baskin Robbins.

Buff · 07-01-2003, 12:22 PM

The train derails after 3 seconds and rolls down the embankment 47 meters at which point it topples into the water and sinks 188.3 meters. Eventually it becomes a coral reef and corrodes in the salt water. Particles of it are dispersed all over the globe by the mighty ocean. In summary, the train never stops traveling.

SilverTab · 07-01-2003, 01:28 PM

Quote:

Originally posted by Porndealer
Battle of the minds
So what's the next question?

Ok an easy one about basic linear algebra...

Are those 3 vectors (in the space R3) linearly dependents or independents

v1 = (3,4,3)
v2 = (2,2,2)
v3 = (5,6,6)

This is by far the course that I hated the most in college...

SilverTab · 07-01-2003, 01:55 PM

Ok a hint...easiest way to solve it:

Put the vector in a 3x3 determinant and solve it....if the answer = 0 then they are linearly dependents, if it's not = 0, then they are independent....

AdultNex · 07-01-2003, 02:00 PM

Quote:

Originally posted by phogirl69
Are you you guys sure this is a high school problem?

Don't ever recall doing anything like that and it looks difficult to me even now, lol.

Perhaps I did take Physics in high school, I really don't remember since it was over 10 years ago and I always HATED math/chemistry/equations.

I was going to go back to college but screw that if I have to do equations and graphs like that!! lol

Yep, it is a Physics question you take in HS. I remember doing so in 10th grade, boy... It was a pain.

SilverTab · 07-01-2003, 02:42 PM

Quote:

Originally posted by SilverTab
Ok a hint...easiest way to solve it:

Put the vector in a 3x3 determinant and solve it....if the answer = 0 then they are linearly dependents, if it's not = 0, then they are independent....

oh well...here we go...

| 3 4 3 |
| 2 2 2 | now we have the 3x3 determinent...
| 5 6 6 |

3 |2 2| = 3(12 - 12) = (3) ( 0 ) = 0
|6 6|

4 |2 2| = -4(12 - 10) = (-4) (2) = -8
|5 6|

3 |2 2| = 3(12 - 10) = (3) (2) = 6
|5 6|

0 - 8 + 6 = -2

so the answer is -2.... the vectors are linearly independent...
Yippi!....I tought someone would've studied a bit of linear algebra...oh well!....

rjstar · 07-01-2003, 04:07 PM

s=ut-0.5at^2
s=40*30+1*900
s=1200+900
s=2100m

v=u+at
0=40+-2t
t=20

So the train travels 2.1km in total but stops after 20 seconds. So the answer is zero, it can go nowhere with it's current accelaration after 30 seconds.

sexeducation · 07-02-2003, 04:50 PM

No my original answer is correct ...

Given ....
Kinetic Energy = 1/2mv^2

and given
Potential Energy = Force X Distance

you get the object currently has maximum Kinetic Energy at it's start veloctiy ...

and minimum kinetic energy when it reaches Zero velocity - which is it's maximum potential energy ...

do the algebra ... and you get the correct answer ...

Wow ...it's been a long time since I did one of these ...
See what becoming a "sex webmaster" does to your brain ...

The only vector I think about now is in - and - out.

Sorry boys ...
My answer/s depending on you definition of "brakes" is correct ...

You do not need graphs and vector analysis or any type of "series" calculation ...

and this really is a high school question ...

What happened to those babes in high school ..

Dad@

BlueDesignStudios · 07-02-2003, 05:41 PM

Quote:

Originally posted by Buff
The train derails after 3 seconds and rolls down the embankment 47 meters at which point it topples into the water and sinks 188.3 meters. Eventually it becomes a coral reef and corrodes in the salt water. Particles of it are dispersed all over the globe by the mighty ocean. In summary, the train never stops traveling.

Your analysis agrees perfectly with my predictions...

06-30-2003, 11:02 PM	#67
High Quality Confirmed User Join Date: Feb 2002 Location: Vegas Posts: 5,741	hey dumb ass, 2m/s/s is an acceleration not a speed. __________________ RecurCash.com - Averaging $38/sale with 60% revshare in the first 4 months alone! Convert your TEEN traffic today @ better than 1:500 guaranteed. ICQ me: 18287590!

06-30-2003, 11:15 PM	#71
eroswebmaster March 1st, 2003 Join Date: Jul 2001 Location: Seat 4 @ Venetian Poker Room Posts: 20,295	What's the deal, your dungeons and dragons groups not show up tonight? __________________ For rent - ICQ 127-027-910 Click here for more details

06-30-2003, 11:17 PM	#75
nosey Talk Hard Join Date: Feb 2003 Posts: 14,413	__________________ \| Domain whois privacy Free \|\| GFY favored Hosting \| $Chaturbate \|\| FpcTraffic FPCPlugs \|\| PlugRush Traffic

06-30-2003, 11:21 PM	#76
BlueDesignStudios Confirmed User Industry Role: Join Date: Feb 2003 Location: Australia Posts: 9,492	Could we have someone write a php script to demonstrate? People here seem to love writing php scripts __________________ Blue Design Studios - Adult Design Specialists! Email me for a free quote: [email protected]

06-30-2003, 11:41 PM	#82
phogirl69 Confirmed User Join Date: Jul 2001 Posts: 3,330	Are you you guys sure this is a high school problem? Don't ever recall doing anything like that and it looks difficult to me even now, lol. Perhaps I did take Physics in high school, I really don't remember since it was over 10 years ago and I always HATED math/chemistry/equations. I was going to go back to college but screw that if I have to do equations and graphs like that!! lol Last edited by phogirl69; 06-30-2003 at 11:43 PM..

06-30-2003, 10:48 PM	#64
buddyjuf Guest Posts: n/a	alright, Im hoping to make this my last post on this thread with a more comprehensible (I hope) graph that I just made I hope this helps you understand better

06-30-2003, 11:01 PM	#65
fr33s3x Confirmed User Join Date: Apr 2003 Posts: 5,064	by the graph we see that after 20 sec, the train will stop, and it doesnt matter if you add an extra 10, the distance wont change. v0=40m/s a=-2m/s^2 t=20s ---------- v=? x=? ---------- v=v0-at (- because the acceleration is negetive) v=40-220 v=40-40 v=0 - the speed after 20 sec -------------------------------------- x=x0+v0t-at^2/2 x=4020-2*20^2/2 x=800-400 x=400 ------------------------------------- the train will travelled 400meter within 20 secs, thats why the train travelled the same distance with 30 sec.. hope i helped.

06-30-2003, 11:14 PM	#69
fr33s3x Confirmed User Join Date: Apr 2003 Posts: 5,064	guys! bdjuf is fucking right! the train will stop after 20 SEC, here is the proof: v=v0-at (- because the acceleration is negetive) ok! we need to know when the speed will be 0, so we will see how much time it takes! so: v0=40m/s a=-2m/s^2 v=0 ----- t=? ---- 0=v0-at 0=40-2t -40=-2t t=20sec the train will stop after 20 sec, thats it! no more argue!

06-30-2003, 11:14 PM	#70
Planet Bob Confirmed User Join Date: May 2001 Posts: 303	380 - thank you for playing

07-01-2003, 12:02 AM	#84
High Quality Confirmed User Join Date: Feb 2002 Location: Vegas Posts: 5,741	Ok here is an approach everyone can follow. Speed (velocity) is miles/hour or distance / time. --> Speed = distance / time --> Distance = speed * time --> Time = speed * distance Initial speed is 40 meters / second. Final speed is 0 meters / second Acceleration is defined as a rate or in distance per second per second. --> Acceleration = meters / second / second (m/(s^s)) Thus the equation is: Speed initial - acceleration(time) = Speed final (m/s) - (m/s/s)(s) = m/s <-- units check out 40m/s - 2m/s/s(s) = 0m/s (note that s is unknown) Solving we get S = 20seconds (deceleration time) It will then stop within 20seconds, the 30s question is irrelevant. Thus, to check the final distance: Distance = speed time OR Distance = Initial speedtime - (accelerationtimetime)/2 (use calculus to prove this...) (meters) = (meters/second)seconds - (meters/second/second)secondsseconds) --> Distance = (40m/s)20seconds - (2m/s/s20*20)/2 = 800meters - 400meters = 400meters distance. __________________ RecurCash.com - Averaging $38/sale with 60% revshare in the first 4 months alone! Convert your TEEN traffic today @ better than 1:500 guaranteed. ICQ me: 18287590!

07-01-2003, 12:05 AM	#85
buddyjuf Guest Posts: n/a	thank you for reconfirming my procedure I await mr.wrongs reply on this thread

07-01-2003, 09:16 AM	#86
sexeducation So Fucking Banned Join Date: Jun 2003 Location: Calgary - Alberta - Canada Posts: 7,315	It depends on your definition of "brakes". To my understanding trains have two different types of "brakes". Friction and applied force by reversing the engines. Therefore, there are two answers depending on the context of the question .... Given ... Acceleration =(FINAL VELOCITY - ORGINAL VELOCITY) divided by TIME We get ... -2MS^2 = (ZERO - 40/MS^2) / TIME 20 seconds of time before the train reaches zero velocity from the relative point in space for the question. Therefore the train went the same distance it would accelerate forward at for 20 seconds ... 2/ms^2 for 20 seconds = 40 meters. However, if the reverse engines are still being applied, which the question appears to be saying ... then you need to add another 10 Seconds of acceleration and we add another 20 Meters. One answer is ... The train traveled 60 Meters from the relative point in time in which the question was posed. Quite simply 30 seconds of acceleration at 2/ms^2. Dad@

07-01-2003, 09:32 AM	#87
sexeducation So Fucking Banned Join Date: Jun 2003 Location: Calgary - Alberta - Canada Posts: 7,315	Nope I screwed up the math ... lol The answer is 20 seconds or 30 seconds of acceleration.

07-01-2003, 09:36 AM	#88
sexeducation So Fucking Banned Join Date: Jun 2003 Location: Calgary - Alberta - Canada Posts: 7,315	See what happens when you do not pay attention in Calculus class ...grrrr lol These questions are fun.

07-01-2003, 10:01 AM	#89
Porndealer Confirmed User Join Date: Jun 2003 Location: Ghetto Posts: 248	Battle of the minds So what's the next question?

07-01-2003, 12:22 PM	#93
Buff GFY Assassin Industry Role: Join Date: May 2003 Posts: 2,993	The train derails after 3 seconds and rolls down the embankment 47 meters at which point it topples into the water and sinks 188.3 meters. Eventually it becomes a coral reef and corrodes in the salt water. Particles of it are dispersed all over the globe by the mighty ocean. In summary, the train never stops traveling.

07-01-2003, 01:55 PM	#95
SilverTab Confirmed User Join Date: Nov 2001 Location: MTL Posts: 5,060	Ok a hint...easiest way to solve it: Put the vector in a 3x3 determinant and solve it....if the answer = 0 then they are linearly dependents, if it's not = 0, then they are independent.... __________________ mmm my sig was too big... no more cool animation but hey still! need php? ICQ: 94586959

07-01-2003, 04:07 PM	#98
rjstar Confirmed User Join Date: Jul 2002 Posts: 353	s=ut-0.5at^2 s=4030+1900 s=1200+900 s=2100m v=u+at 0=40+-2t t=20 So the train travels 2.1km in total but stops after 20 seconds. So the answer is zero, it can go nowhere with it's current accelaration after 30 seconds.

07-02-2003, 04:50 PM	#99
sexeducation So Fucking Banned Join Date: Jun 2003 Location: Calgary - Alberta - Canada Posts: 7,315	No my original answer is correct ... Given .... Kinetic Energy = 1/2mv^2 and given Potential Energy = Force X Distance you get the object currently has maximum Kinetic Energy at it's start veloctiy ... and minimum kinetic energy when it reaches Zero velocity - which is it's maximum potential energy ... do the algebra ... and you get the correct answer ... Wow ...it's been a long time since I did one of these ... See what becoming a "sex webmaster" does to your brain ... The only vector I think about now is in - and - out. Sorry boys ... My answer/s depending on you definition of "brakes" is correct ... You do not need graphs and vector analysis or any type of "series" calculation ... and this really is a high school question ... What happened to those babes in high school .. Dad@