Relative Rate or Speed

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Relative Rate or Speed

by shanice » Thu Aug 23, 2012 10:03 pm
I came across a website provided by VERITAS PREP on explanation about relative speed. I still can't understand it. I really need some clear explanation about this. Please refer to the following example provided by Veritas Prep:-


Example:


Say, there are two people, A and B.

Case 1: A is standing still and B is moving at a speed of 5 mph.

What is B's speed relative to A? B is moving away from A at a rate of 5 miles every hour. So B's speed relative to A is 5 mph. What is A's speed relative to B? Is it 0? No! A's speed relative to Earth is 0. A's speed relative to B is 5 mph since distance between A and B is increasing at a rate of 5 mph. Confused? When we say 'relative to B', we assume that B is stationary. Since distance between the two is increasing at a rate of 5 mph, we say A's speed relative to B is 5 mph.

My questions:-
1)Relative means "in comparison" right? So, the 1st sentence is asking "What is B's speed in
comparison(in relative) to A", right?
2)6th sentence - A's speed relative to Earth is 0. What does it mean?I know A's not moving so it
should be 0 but unfortunately it's wrong. It say's in comparison to Earth is 0. I
don't get it.
3)7th sentence - "A's speed relative to B is 5 mph since distance between A and B is increasing at a
rate of 5 mph". A's speed to B is 0 right because he's not moving, why it is
increasing even though,B is considered stationary(not moving).I'm so confused here.
Please give me an example that could help me out.

I'm pretty weak in this relative topic. Please help.

Thank you.

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by Ian Stewart » Fri Aug 24, 2012 2:21 am
I find that a slightly confusing way to think about things, and you can think about this concept differently if you like. It's on questions where you have two things moving simultaneously where these concepts are relevant, so in a question like the following:

Two runners begin running towards each other at the same time from opposite ends of a 200 meter track. If one runner runs at a constant speed of 9 meters per second, and the other runner runs at a constant speed of 7 meters per second, how long will it take for them to meet?

We can sketch what is happening here:

---9 m/s--->................200 m....................<---7 m/s---

Now, you could use some 'relative speed' concept to answer this question, but instead, you can just ask "how does the distance separating these two runners change each second?" In the first second, the distance between them falls by 9 meters on the left, and by a further 7 meters on the right, so it will fall by a total of 9 + 7 = 16 meters. That happens each second, so the distance between them is falling at a speed of 16 meters per second. We need the distance between them to fall by 200 meters, so using the t = d/s formula, we find that it will take t = 200/16 = 12.5 seconds.

You can do the same when your runners run in the same direction:

Two sprinters begin a race at the same time, running in the same direction. If the faster sprinter runs at a speed of 10 m/s, and the slower sprinter runs at a speed of 8.5 m/s, how long will it be before they are 9 meters apart?

Again we can sketch what is happening:

--------10 m/s--------->

-------8.5 m/s---->

Now in the first second, the distance between them grows by the difference in the speeds, so by 10 - 8.5 = 1.5 meters. So the distance between them is growing at a speed of 1.5 meters per second, and we want the distance to grow to 9 meters. Using t = d/s, we find it will take t = 9/1.5 = 6 seconds.

There are a few other variations, but if you just ask on any of these questions "what happens to the distance between my objects in the first second/minute/hour?" the answer to that question gives you a speed, the speed at which the distance between your objects is changing. You can always use that speed to answer your question.
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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by shanice » Sat Aug 25, 2012 10:05 pm
Hi Ian,

Thank you so much for taking your time to explain about this topic.

I understand the second scenario very well but I'm having a problem with the first one. I have one doubt about scenario one from the following sentence:-

1)In the first second, the distance between them falls by 9 meters on the left, and by a further 7 meters on the right, so it will fall by a total of 9 + 7 = 16 meters.

My understanding - Since both of them are running towards each other at the same time from opposite ends, I can imagine that the distance between them gets closer(shrinking or nearer). I get that part. But why do we have to add their rates? I'm confused about that.

Would you mind explaining it to me again in detail for this part only with easy examples? I'm sorry if my question sounds silly but I really want to understand it instead of memorizing it.

Thank you in advance, Ian.

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by shanice » Sun Aug 26, 2012 6:06 am
Can anyone please help me on this?

Really appreciate it.

Thank you.

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by Ian Stewart » Sun Aug 26, 2012 1:25 pm
shanice wrote:
My understanding - Since both of them are running towards each other at the same time from opposite ends, I can imagine that the distance between them gets closer(shrinking or nearer). I get that part. But why do we have to add their rates? I'm confused about that.
If they are running towards each other, one at 9 meters/second, the other at 7 meters/second, then in 1 second the first runner runs 9 meters, and the second runner runs 7 meters. If they began 200 meters apart, then in 1 second they have reduced the distance between them by the total distance they ran, and they ran a total of 9+7 = 16 meters. That's why we add their speeds - that tells us how much distance they are covering together each second.
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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by shanice » Sun Aug 26, 2012 8:20 pm
A million thanks to you, Ian. I understand it now. I can't wait to try some questions.

Thanks, again.