4 hours 30 minutes? 30 x 2 = 60 , 60km/h is double speed , 6 hours is now 3 hours because you go double speed? and john's ride in 45km/h should then be 4 hours and 30minutes.
i am probably wrong, just made it as i count :]
_________________ <<banned from SRF for the sale of digital items for cash. -SG>>
u got the 4 hours right but not the 30 mins i dnt get how u did it though
sry my english aint the best ;P
first of all i said > 30km/h x 2 = 60km/h < and the trip was 6hours when john was driving in 30km/h and if u double 30hm/h as i did between the *> <* that will make that the ride will be 3 hours because john is going the double speed (not 30km/h , 60hm/h). BUT he was going 45km/h so im just doing a calculate thingy or w/e its called here . 30 * 2 = 60 , imagine 4 balls tied to eachother, and each ball is 15km/h >all put together is 60km/h< , then take away one ball, and you will have 45km/h and 45km/h = 4 hours and 30minutes, (remember that 60km/h = 3hours and 30km/h = 6hours.
hope that helped
_________________ <<banned from SRF for the sale of digital items for cash. -SG>>
yes something to do with einstein theory of relativity and frame of reference? u dont know about that?
lol. That is a very simple word problem with a very simple equation. Einstein's theory of Relativity has nothing to do with it.
"truth is not a democracy" just because you choose to deny it doesn't make it true.
yes there is.. whenever theres movement (motion) then relativity comes into play.
you are the one thats denying it even though i solve the problem for x.
and you havent answer my post about the displacement thingy since you claim to know the answer.
x = displacement, in some equations displacement (x) is replaced with d (distance). very simple equation, very simple answer. I'm not denying anything, I used the traditional time equation ~ and in order to find time I needed to find x (displacement).
now don't you have one unknown you can solve for the other (t).
t = x/v v = 45km/h x = 180km t = (180km)/(45km/h) t = 4
very simple
what did u just do? y would u do it the long and wrong way? i dont see why you would make something so easy... so complicated. my way of solving makes so much more sense.
now don't you have one unknown you can solve for the other (t).
t = x/v v = 45km/h x = 180km t = (180km)/(45km/h) t = 4
very simple
what did u just do? y would u do it the long and wrong way? i dont see why you would make something so easy... so complicated. my way of solving makes so much more sense.
but anyway u are wrong my friend.
when distance traveled is a constant and velocity increase time will decrease. The question isn't complex, its either a 1st year Physics question or a Algebra I question. Theoretical physics has a place, this word problem is not where it is applied. fyi, you're wrong not me buddy.
so, let me get this straight. If you're traveling the same distance at a greater velocity it should take you more time?
yes something to do with einstein theory of relativity and frame of reference. u dont know about that?
i already solved it
Quote:
6hours/30kmh = x/45kmh
solving for x...
45*6 = 30x x= 270/30 x = 9 hours
Greetings !G4!, The question: A journey takes 6 hours if John travels 30km/h. How long will it take him if he travels at 45km/h?
Let's think about it for a second before we "solve" it.
"How long?" equals "X"
Spoiler!
Multiplication Property of Equality means: If a = b, then a(c) = b(c) Division Property of Equality means: If a = b, then a/c = b/c where c is not equal to 0
Make a equivalence: '6 is to 30' is the same as -or- equivalent to 'X is to 45'
Use the Multiplication Property of Equality:
6 × 30 = 45 × X
Solve for one side of the equation: 6 × 30 = 180
Use the Division Property of Equality
180 = 45X <~~~~~~~ Divide both sides by 45 to solve for "X"
180 divided by 45 = 4 = X
X = 4
!G4! Got the answer right, but didn't show the "proof" StealMySoda wins!
About Einstein: he said, "The only real valuable thing is intuition."
Lol @ Kazaxat and Barotix. Yeah, pretty much every poster except for Kazaxat is correct. You want to take v1*t1 / v2 = t2, where v1 = first velocity, t1 = first time, and v2 is the second velocity (45 k/hr)
30 * 6 / 45 = 180 / 45 = 4
_________________ McCain, he (Barack Obama) said, will soon "be accusing me of being a secret communist because I shared my toys in kindergarten."
Kazaxat, don't try to sound smart about it, the answer is clearly 4 hours. There is no crap about relativity involved in a simple algebra problem. The other posters already solved it many times over correctly without adding random things that don't belong. They correctly found the distance, which is exactly 180 km. Divide by 45 km/h and get 4 hours. D = RT.
By the way, you tried to set up a proportion. That doesn't work with speed/time/distance problems. They are directly related by the reciprocal property. If one thing (like the rate) is changed, then another thing (like the time it takes to get there) is inversely and equally changed. Not sure where relativity comes in here.
Faster rate = Less time to get there Slower rate = More time to get there
It makes sense that a 45 km/h journey would take less time than a 30 km/h journey. It's 1.5x the speed. So the time for a 30 km/h journey should be (inversely) 1.5x the time of a 45 km/h journey. 6 = 1.5 x 4.
_________________ Adv4nc3chao: Level 89 Hybrid Warlock/Cleric Quit since Jan. 2009
That one can absolutely conclude about motion (and gravities) of planets and stars by observing cannonballs.
How can Einstein's Special Principle of Relativity rightly apply?
Spoiler!
The basal principle of all uniform motion rightly applies if the trains were on large space ships while they were approaching the speed of light in a vacuum ("C") with differing velocities. Velocity is defined in terms of direction and speed but even if speed and acceleration were equal, as long as the two ships had different vectors 'relativity' and Einstein's "Law of Optics" would rightly apply.
Einstein states the Law of Inertia as follows:
"A body removed sufficiently far from other bodies continues in a state of rest or of uniform motion in a straight line."
Albert Einstein (1879–1955). Relativity: The Special and General Theory. 1920. (p. 13)
This law is important because the Newtonian Mechanics do not hold for a system where this law does not hold. Since we can rightly assume that John and the train or trains in question were on earth, at the time of the test, your answer (X=4) is correct. But nevermind me, I'm trying to get my post count to 100.
but oh well i will start looking at the problem again tomorrow with a fresh mind.
That's correct, you would be right... ...If only the problem was not related to rates and distance.
Proportions are for other types of problems that don't have a reciprocal relationship. For example, if you had a problem like this:
"Bob worked 6 hours and made $30. How long would he have worked if he made $45?"
Setting up a proportion would in this kind of problem since money gained depends directly on hours worked. 6/x = 30/45
Therefore, you do 45 x 6 / 30 = 9 hours. See? The reciprocal property does not apply to fixed rates. The hours worked and money gained do not influence each other like rate and distance do. Do you understand?
More time worked = More money Less time worked = Less money 1.5x time worked = 1.5x money
Whereas, like I said before,
Quote:
Faster rate = Less time to get there Slower rate = More time to get there
They are different things. You were actually partly correct but unfortunately mislead. Speed and distance problems have their own little equation; D = RT.
Kazaxat wrote:
6/30km = x/45km
so 9 hours.
@Grandpa: mmk
_________________ Adv4nc3chao: Level 89 Hybrid Warlock/Cleric Quit since Jan. 2009
Last edited by Advancechao on Sat Apr 12, 2008 6:54 am, edited 4 times in total.
One -CAN- express mathematical problems like this in terms of fractions or ratios (or percentages for that matter). -BUT- it is not the considered the best way because
It will have compound fractions if we try because
Rates (like kph or $ per hr.) are terms divided by hours
-AND- fractions are expressions of division
-AND- Reciprocals are defined as two numbers that multiply to equal 1 (for example, 3/4 and 4/3 are reciprocals --> 3/4 × 4/3 = 12/12 = 1).
-SO IF- we insist on using fractions when we set up the 'equality'... We must be sure to keep Like Terms on the same side of the equation then solve for X...
(The only thing we CAN'T 'properly' do is divide by zero)
HINT: Set the reciprocal up so that the correct answer equals X = 4 = Trip One / Trip 2 × Trip 2 / Trip One = 1 (it gives me a headache to think of it)
Spoiler!
Advancechao did it for you (for the amount of $ per hour question):
Quote:
Setting up a proportion would work here. 6/X = 30/45
But to use the reciprocals (if you don't mind compound fractions) be sure to keep Like Terms on the same side of the equation: Reciprocal: (( 6hrs) / (Xhrs) = (30 dollars per hour / 45 dollars per hour )) × (($30 / $45 ) / (6hrs) / (Xhrs)) = 1 / 1 == 1
(I'm getting silly here)
For the "easy way" and "Trains" problem the equality is: 6 is to 30 as 45 is to X. Use multiplication (it's easier): 6 × 30 = 45 × X
To use division the equality becomes: 30 / 45 = X / 6
When directly comparing the problems of Bob and John... It's generally better to leave Advanced Ratios till later but; Many times we do use ratios to solve word problems.
So... Question: Can it be said that for 100% of the Time (t), while comparing ratios to fractions, we are using proportions -AND- when we use non-reciprocating proportions we are comparing inequalities
HERE ~~> Have a crazed cookie: Teachers who tell us to avoid Division and use Multiplication when possible have eaten too many of these! I enjoy finding similarities (especially with word problems) so: It can be shown that ( 9 ∙ 0.75 ) = 6 and ( 0.75 = ¾ ) -BUT- ( ¾ ≠ 40/50 )
For the John & the "trains" ratio use 30:45 -or- 45:30 (which is the same as 6/9 -or- 9/6 (reciprocals) but ≠ 4/5).
For Bob & the '$ per hour' ratio use 30:40 (which is the same as 6/8 -and- ¾ but ≠ 4/5 )
Comparing both problems -CAN- be done, but it is similar to asking, "-What if- we change Bob's payrate to introduce Time (t) from $5.00 to $4.00 / hour) -and- how long he worked (term)? This introduces time (t) as a direct factor as is commonly done in economic equations but it is too complicated for 'Basic Algebra' and similar to introducing vectors to the distance equation without Einstein's Relativity laws (non-sequitur).
Comparing numbers that when multiplied ≠ 1 (non-reciprocal) is like comparing apples to oranges.
Such things (when reduced) are similar to saying 4 = 5.
Although it is true that John took 4 hours on the second trip -AND- Bob was paid $5.00 / hour comparing the variance between Time regarding Bob's (t) per dollar and John's (t) per trip is of little use. BTW this @ Advancechao: mmk /// wat is 'mmk'?
It's much better to use the simpler (Basic Algebra) Multiplication and Division Properties of Equality seen > HERE <
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So... Question: Can it be said that for 100% of the Time (t), while comparing ratios to fractions, we are using proportions 
