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Shrinking and Expanding Gaps
Shrinking expanding gaps. Some motion based word problems involve two travelers moving toward or away from each other. Some very basic rules make these problems much easier. So, first of all, let’s consider the case where two travelers are moving in opposite directions.
When the two travelers are moving in opposite directions, either toward each other, or away from each other, we always add the speeds. If the two travelers are approaching each other, the sum of the speeds is the speed at which the gap is shrinking, or closing. If the two travelers are moving away from each other, the sum of the speeds is the speed at which the gap is expanding.
When two travelers are moving in the same direction, we always subtract the speeds. Bigger minus smaller. If the faster traveler is in front, then the difference in speeds is the speed at which the gap is expanding. If the slower traveler is in front, then the difference in speeds is the speed at which the gap is shrinking, or closing.
In a problem in which a gap is obviously shrinking or expanding, because we have two different cars and two different speeds, of course we could set up a D equals RT equation for each traveler, for each vehicle. But sometimes it saves time to set up a D equals RT for the gap itself, the distance of the gap, the speed of the gap, and the time that it takes for the gap to expand, or to cover that distance.
If we figure out the speed at which the gap is shrinking or expanding, we can relate the two individual speeds Bb simple addition or subtraction, depending on how, depending on whether they’re moving toward each other or away from each other, what the configuration is. Here’s a practice problem. Pause the video, and then we’ll talk about this.
Okay. This would be a relatively hard problem if we set it up in terms of assigning variables for the car, variables for the truck, etc., etc. That could be a very hard way to do it. I think it’s much easier, here, to think about the gap, because clearly, at 3:00 P.M. there’s a thirty mile gap between them.
And at 4:30, the car passes the truck, so there’s no gap. So the gap went from thirty miles to zero in that time period, 3:00 to 4:30, which is an hour and a half. So we could actually figure out a speed for that gap. R equals D over T. Distance is 30.
Time is an hour and a half. 30 divided by an hour and a half, well, let’s see. We could say that that is, wee could multiply top and bottom by 10, and so it would be 300 over 15. Of course, 30 divided by 15 is 2, so 300 divided by 15 is 20. And so that is the rate of the gap.
The gap was changing at 20 miles an hour. Well, this is going to be the distance, the difference between two vehicles because they’re moving in the same direction. The car was behind the truck, moving toward the truck, and passing the truck. So they’re moving in the same direction, and 20 is the difference in their speeds. And the slower one is the truck, the faster one is the car, because of course the car passes the truck.
So the car must be moving at 20 miles an hour faster than the truck. Well, the truck is moving at 50 miles an hour. So it means that the car has to be moving at 70 miles an hour. Here’s another practice problem. Pause the video, and then we’ll talk about this. Okay.
Car X and Y are traveling from A to B on the same route at constant speeds. So, they are traveling in the same direction. Car X is initially behind car Y, but car X’s speed is 1.5 times car Y’s speed. All right. Well, that’s interesting, that’s kind of an unusual requirement. Car X passes Car Y at 1:30.
So, at that moment, when one car passes the other, they’re at the same place, there’s no gap between them. And then later on, at 3:15, car X reaches the destination, reaches point B. And at that moment, the moment when car X reaches point B, car Y is still 30 miles away from B. So it means that, at that point, there’s a separation between them of 35 miles.
So we want to know, what is the speed of car X, which is the faster car? So, just so you can visualize this, starts out with this fast car X is behind the slower car, and it’s catching up at 1:30 P.M. Car X passes car Y, so for a split second they are at exactly the same place at 1:30. And then by 3:00 P.M., of course, car Y has moved ahead a little bit, but car X has gotten all the way to the destination B, and at that point a 35 mile gap between the two cars.
So first of all, let’s think about this time interval from 1:30 to 3:15. Well 1:30 to 2:00 is half an hour, 2:00 to 3:00 is an hour, 3:00 to 3:15 is a quarter of an hour. Adding all those up we get 1 plus three quarters, or seven quarters. And so, this time period is 7 quarters of an hour. So that distance of 35 miles opened up in 7 quarters of an hour.
And so the gap of 35 miles, that’s the gap distance, and then the gap time is 7 over 4. And so this is the same as 35 over 1 times 4 over 7. Cancel the 7’s, we get 5 times 4, which is 20 miles an hour. And so that’s the speed associated with the gap. Because the two vehicles were moving in the same direction, that is the difference in speeds.
We know that the fast car minus the slow car is 20 miles an hour. So, from the prompt, we know that X is 1.5 times Y. That’s the unusual equation given as part of the prompt. And we just figured out also that the difference between them is 20, X minus Y is 20.
That’s the difference. And so now, we’re going to plug in X from the first equation. The first equation is solved for X, so plug that into the second equation. We get 1.5Y minus Y equals 20. So 1.25 minus 1 is 0.25, or a quarter. And then multiple both sides by 4 and we get Y equals 80.
So 80 the speed of car Y, and car X is going 20 miles an hour faster than that. Or, ultimately, take 80 and multiply by 1.25. Either way, what we’re gonna get is the speed of x which is 100 miles an hour. So that is the speed of car x. Here’s another practice problem. Pause the video, and then we’ll talk about this.
So this is interesting. This one, the distances are all given, and the speeds are given, but what we don’t know is the time. We’re looking for a clock time. Very interesting. So car P is moving at 49 miles an hour.
And car Q is moving at 61 miles an hour, and they are approaching each other. They’re moving in opposite directions. At 2 P.M. they’re approaching each other and are 121 miles apart. Okay. So then they approach, and then they pass each other. So at that point the gap between them is 0.
And then we want to know at what time are they moving away from each other and are 44 miles apart. Well, this is very interesting. So first of all notice, this is the basic scenario we have here. So at 2 P.M. they’re 120 miles apart, and they’re approaching each other. So then that gap closes to 0 when they drive right past each other, they pass each other at some unknown time.
And then we want to know the clock time when they are now 44 miles apart, but going in opposite directions. So, because they’re moving in opposite directions we can add the speeds. So the gap is changing at 110 miles an hour. And really, there’s two gaps here. There’s the first gap when they were approaching, and that gap is 121 mile gap, and then there’s a second gap that’s receding.
And so, each one of these would get divided by the speed, and that would be the time for that particular gap, and then we’re gonna add them together. Well, it turns out, we could just add those two numbers together right away, and then divide. And that’s a little bit easier. Rather than adding fractions, we’ll just add these two numbers, 121 plus 44 together.
That’s 165. So that’s really a combination of two gaps there, together, all changing at 110mph because it doesn’t matter whether they’re approaching or receding, we still add the velocities. So 110mph, 165 miles divided by 110 miles per hour, turns out that both of those numbers are divisible by 11.
So cancel the 11’s, and we get 15 over ten, or 1.5. So 1.5 hours. So 1.5 hours from the start. The start was at 2 P.M. One hour after that is 3 P.M. And another half hour is 3:30. And that’s the answer.
Two things are going in opposite directions means we add. Going in the same direction means we subtract. We have to think about directions to determine whether the gaps are expanding or shrinking. And solving a D equals RT for the gap itself can enormously simplify such problems.
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