The express train starts off later. See post #18.
Bob
]]>hi Denominator
Let the distance from Southampton to London be D, and the distance to the overtaking point be d.
Let the time in minutes from 9.17 to the overtaking point be t
The goods train takes 165 minutes
The express train takes 100 minutesSo the gradient (velocity in miles / min) for the goods train is
and for the express train
You can eliminate both D and d by dividing these equations:
and then solve for t.
Bob
How did you get that 39? Sorry, couldn't comprehend.
]]>Bob
]]>Your welcome!
]]>Thank you bob and bobby and anonim!
]]>Let the distance from Southampton to London be D, and the distance to the overtaking point be d.
Let the time in minutes from 9.17 to the overtaking point be t
The goods train takes 165 minutes
The express train takes 100 minutes
So the gradient (velocity in miles / min) for the goods train is
and for the express train
You can eliminate both D and d by dividing these equations:
and then solve for t.
Bob
]]>Okay, yes you are right, I did not see that.
]]>
Original equation:
Different equation:
]]>Unless your saying the roots of the original equation are 1/2 and √19 / 2
Which I don't think is right
Isn't 5 also an answer?
]]>Yeah, after using your methods, I get 5/4 as well
I tried doing something different:
Let the solutions of the different equation be X and 2
The solutions of the original equation is 1/X and 1/2 (after playing around, I found this rule)
So 1/X=X
Which means X=1 or -1
Sum of squares (1)²+(1/2)²=5/4
Btw even though 2x²-5x+2 works, the question says 2 different equations, 1 original and the other with the coefficients swapped... this stays the same.
Thanks a lot guys!
As for Q2
Can you tell me what equations you formed, because I couldn't make some..
trains:
For questions like this I like to sketch a distance time graph so i get a good picture in my head of what is happening. see below.
You have enough data to work out the gradients for the two lines. And you know that at the moment of passing the 't' values and the 'd' values are equal.
So you could make two equations for d/t based on the gradients and solve for d and t.
Bob
]]>You also should have posted a new thread instead of tacking on to an active one.
]]>