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DOJO
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#1
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The brakes of a train can produce a retartdation of 1.7ms-2. If the train is travelling at 100kmh-1 and applies its brakes, what distance does it travel before stopping? If the driver applies the brakes 15m too late to stop at a station with what speed is the train travelling when it passes through that station?

I got hte first part correct:

U= 100kmh-1

after converted

27.8ms -1

a= -1.7ms-2

Equte them variables into the formulae to get:

V^2=U^2+2as
0=27.8^2+2(-1.7)s
s=772.84/3.4 = 227m

How do you do part B? Thanks!
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Fermat
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The train needs 227 m to stop, So it puts the brakes on 15m late which means it will travel 15m past the station. What you have to find is the speed of the train when it reaches the station which will be when it has travelled 227 - 15 = 212 m.
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magiccarpet
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if what fermat says is right, this should be:
v=?
u=100km/h
t=15
a=-1.7ms-2

v=u+at
=100000+(15*60*-1.7)
=98.47kms-1 i think im trying and i could be wrong
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DOJO
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(Original post by magiccarpet)
if what fermat says is right, this should be:
v=?
u=100km/h
t=15
a=-1.7ms-2

v=u+at
=100000+(15*60*-1.7)
=98.47kms-1 i think im trying and i could be wrong
The answer is 7.14ms-1 :elefant:
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magiccarpet
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crap im in the library gotta go soon cos its closing. when do you need to know for? cos i really wana know how to do it, i can do it tonite and send u the ansewr tomoro bout 10am?
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DOJO
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(Original post by magiccarpet)
crap im in the library gotta go soon cos its closing. when do you need to know for? cos i really wana know how to do it, i can do it tonite and send u the ansewr tomoro bout 10am?
I have moved on in the chapter , I am not too concerned if I cant do it. But it would be nice if someone did solve it 10am tommorrow is fine.
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magiccarpet
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actually what fermat says is wrong. just quickly, the 15 mins too late to stop at the station means 15 mins of the total stopping time not 15 mins=15metres at 100km. wotk out total stopping time using u=0 v=100 a=-1.7 then subtract 15 mins from it should then work it out using t=your answer,u=? v=100 a=1.7. i got 10 mins gonna try now but im sure this is how
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DOJO
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(Original post by magiccarpet)
actually what fermat says is wrong. just quickly, the 15 mins too late to stop at the station means 15 mins of the total stopping time not 15 mins=15metres at 100km. wotk out total stopping time using u=0 v=100 a=-1.7 then subtract 15 mins from it should then work it out using t=your answer,u=? v=100 a=1.7. i got 10 mins gonna try now but im sure this is how
Well good luck, I am not sure what formulae to use
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magiccarpet
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ok well im off now. put your t into v=u+at with u=? v=100 a=1.7
i think this is how but i cant do it now cos i gotta go and the library opens at 10am 2mo so if i can do it i send it to you then via pm
good night!
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DOJO
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(Original post by magiccarpet)
ok well im off now. put your t into v=u+at with u=? v=100 a=1.7
i think this is how but i cant do it now cos i gotta go and the library opens at 10am 2mo so if i can do it i send it to you then via pm
good night!
Thanks for the help goodnight!
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Juno
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I can get 7.21m/s!

v^2=u^2+2as

v we want to find
u=100km/h
a=-1.7
s=227-15=212
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Fermat
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(Original post by juno_the)
I can get 7.21m/s!

v^2=u^2+2as

v we want to find
u=100km/h
a=-1.7
s=227-15=212
You're close!
That's the right formula as well.

v² = u² -2ds

v = ?
u = 100km/h = 27.78 m/s
d = 1.7 m/s²
s = 227 - 15 = 212 m

v² = (27.78)² - 2*1.7*212
v² = 771.73 - 720.8
v² = 50.93
v = 7.136
v = 7.14 m/s
=========
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username9816
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(Original post by DOJO)
The brakes of a train can produce a retartdation of 1.7ms-2.

1.) If the train is travelling at 100 kmh-1 and applies its brakes, what distance does it travel before stopping?

2.) If the driver applies the brakes 15m too late to stop at a station, with what speed is the train travelling when it passes through that station?
1.) From applying the brakes to stopping:
a = -1.7, u = (100 * 10^3)/(60^2) = 250/9 ms^-1, v = 0.
v^2 = u^2 + 2as
---> 0 = (250/9)^2 - 2(1.7)s
---> 3.4s = (250^2)/81
---> s = (250^2)/(81 * 3.4)
---> Distance Travelled (To Stop) = 226.9 m

2.) From passing through the station to stopping:
a = -1.7, s = 15, v = 0.
v^2 = u^2 + 2as
---> 0 = u^2 - 2(1.7)(15)
---> u^2 = 51
---> Speed (Passing Through Station) = 7.14 ms^-1
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DOJO
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(Original post by Nima)
1.) From applying the brakes to stopping:
a = -1.7, u = (100 * 10^3)/(60^2) = 250/9 ms^-1, v = 0.
v^2 = u^2 + 2as
---> 0 = (250/9)^2 - 2(1.7)s
---> 3.42s = (250^2)/81
---> s = (250^2)/(81 * 3.42)
---> Distance Travelled (To Stop) = 225.6 m

2.) From passing through the station to stopping:
a = -1.7, s = 15, v = 0.
v^2 = u^2 + 2as
---> 0 = u^2 - 2(1.7)(15)
---> u^2 = 51
---> Speed (Passing Through Station) = 7.14 ms^-1
First part is 227m. Second part is correct
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DOJO
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#15
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(Original post by Nima)
You're wrong, the first part is 225.6 m.

You rounded quantities during the question, included initial velocity, and thus got a slightly less accurate answer.

I used 250/9 ms^-1, and went from there, which was more accurate.
Oh I see, well in the back of the book it is 227ms for that part of the question.
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Fermat
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(Original post by Nima)
---> 0 = (250/9)^2 - 2(1.7)s
---> 3.42s = (250^2)/81
---> s = (250^2)/(81 * 3.42).
ahem.. 2*1.7 = 3.4!
:p:
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Juno
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(Original post by Fermat)
You're close!
That's the right formula as well.

v² = u² -2ds

v = ?
u = 100km/h = 27.78 m/s
d = 1.7 m/s²
s = 227 - 15 = 212 m

v² = (27.78)² - 2*1.7*212
v² = 771.73 - 720.8
v² = 50.93
v = 7.136
v = 7.14 m/s
=========
I do stats now. That's my excuse
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magiccarpet
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#18
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hey dojo sorry i overslept this morning but i had a go. i can only get the right answer if i take 15m to mean 15 metres instead of 15 minutes as the total stopping time is about 16 seconds. but looking above i see the guys have already solved it. oh well
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DOJO
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(Original post by magiccarpet)
hey dojo sorry i overslept this morning but i had a go. i can only get the right answer if i take 15m to mean 15 metres instead of 15 minutes as the total stopping time is about 16 seconds. but looking above i see the guys have already solved it. oh well
No problem!
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