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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]
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SaraLotfy wrote:
I don't know how to solve these distance/rate question in the two "meeting" at some point in the middle. Some questions ask at what time would they meet and others ask about the distance at which they would meet.

Can someone please clarify the equation I should use. I know I'm supposed to subtract rates, but can someone please give me a quick run down over the procedures.

Thanks


Remember 2 things :

I.Two objects moving in the same direction,subtract their speeds, call it \(V_{sd}\)

II.Two objects moving in opposite direction, add their speeds, call it \(V_{od}\)

The distance between them be fixed as D. Thus, as we know that Distance = Speed*Time, for the first case, we have Time taken to meet = \(\frac{D}{V_{sd}}\)

Similarly, for the second case, we have Time = \(\frac{D}{V_{od}}\)

Now, if you want the distance at which they meet, just multiply the time calculated above, fix the reference from which object do you want to calculate the distance, and multiply this time with the speed of that object.

Also, you can refer to this post : cars-p-and-q-started-simultaneously-from-opposite-ends-of-a-159355.html#p1264794


Hope this helps,
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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]
mau5 wrote:
SaraLotfy wrote:
I don't know how to solve these distance/rate question in the two "meeting" at some point in the middle. Some questions ask at what time would they meet and others ask about the distance at which they would meet.

Can someone please clarify the equation I should use. I know I'm supposed to subtract rates, but can someone please give me a quick run down over the procedures.

Thanks


Remember 2 things :

I.Two objects moving in the same direction,subtract their speeds, call it \(V_{sd}\)

II.Two objects moving in opposite direction, add their speeds, call it \(V_{od}\)

The distance between them be fixed as D. Thus, as we know that Distance = Speed*Time, for the first case, we have Time taken to meet = \(\frac{D}{V_{sd}}\)

Similarly, for the second case, we have Time = \(\frac{D}{V_{od}}\)

Now, if you want the distance at which they meet, just multiply the time calculated above, fix the reference from which object do you want to calculate the distance, and multiply this time with the speed of that object.

Also, you can refer to this post : cars-p-and-q-started-simultaneously-from-opposite-ends-of-a-159355.html#p1264794


Hope this helps,


Thank you, this is exactly what I needed :)
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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]
honchos wrote:
Station Q is to the East of Station T. At 12 noon, a train starts from Station Q and travels at a constant speed of x mph towards Station T. At 12 noon of the same day, another train starts from Station T and travels at a constant speed of y mph towards Station Q. At what time will the trains meet?

(1) y = 4x/3
(2) x = 100 mph


Oh I see, they got me on this one.

E for sure

Cheers!
J :)
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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]
seems easy, but it is actually a good question.

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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]
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