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

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 :1orglaugh

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.

:thumbsup

thank you for reconfirming my procedure

I await mr.wrongs reply on this thread :thumbsup

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@

Nope I screwed up the math ... lol

The answer is 20 seconds or 30 seconds of acceleration.

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

lol

These questions are fun.

Battle of the minds :1orglaugh
So what's the next question?

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.:1orglaugh

Good job!

thanx dude, it was fun debating you :thumbsup

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

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.

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)

:Graucho

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

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....

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

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!....

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.

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@

Your analysis agrees perfectly with my predictions... :Graucho

DID - the train actually exist? ...lol


Luc Duboi - did you find out the correct answer?