Jack and Christina are standing 210 feet apart on a level

Well done, guys.
I see that you are advanced already in this kind of exercises. Somehow you kind of leap steps, at least just for me, a beginner.
What I have learned is that is "easier" to combine the speeds of Jack and Christina to find out the time:
3 feet per second + 4 feet per second = 7 feet per second
Then, divide: 210/7 = 30 seconds.
Then as you did, multiply for the speed of Lindy: 30*10 = 300

Hey Cano!
I would like to tell you something!
First, the solution of redjam is much better that mine to understand how it works. But I did mine because I did some questions like this one and I am telling you, the answer is never too complicated! That's why on those kind of questions, instead of thinkig of the equation I should ind, I am looking for an easy to solve.

In this particular case, it seems really complicated! So first, when do they meet.
Okay, how much feet in 30 seconds for J and C? Did they meet? Not yeat ! So how about in 1 minute ? etc etc... usually it works pretty fastly...
And then, you know when they meet, you just have to compute he the distance the dog run!! And don't think about: the dog has to do a U-turn, it will take time, etc etc.... imagine the dog is running as much as time as J and C need to meet.

Hope I get myself clear

Good question (a KUDOS from me for that). And yes, I have seen such a question before.

Thought it was a good question as well. The math isn't hard its more in the framing of the question. The reader could easily be confused and start to try and calculate each there and back by the dog instead of quickly determining the time the two people would meet and then multiplying out the time the dog would be constantly running.

Nice question on relative speed. Easy 300 using relative concepts liek cano

Initially I had a different approach!

we know J & C will meet in 210/7 = 30 sec, J will travel 120 mtrs and C will travel 90 mtrs

At start L moves towards J and they have relative speed of 10+4 and they meet after 210/14=15 sec

In 15 sec L traveled 150 mtrs --- 1

Now in 15 sec C traveled 45 meters and J traveled 60 meters So now distance between J & C is 105 mtrs

when L moves towards C they have relative speed 10+3=13

so now L & C meet in 105/13= 8 sec aprrox and in this time L traveled 80 meters --- 2

C now is at 23*3=69 meters she covers remaining 90-69 meters in 7 sec
L covers 70 meters in 7 sec -- 3

So total dist covered 150+80+70=300 mtrs

redjam,

Shouldn't we consider relative speed of Lindy with Christina and Jack? Sometimes it gets added up and sometimes wen Lindy returns back to Christina we may have to subtract the speeds of Lindy and Christina? Please explain

Jack and Christina are standing 210 feet apart on a level surface. Their dog, Lindy, is standing next to Christina. At the same time, they all begin moving toward each other. Jack walks in a straight line toward Christina at a constant speed of 3 feet per second and Christina walks in a straight line toward Jack at a constant speed of 4 feet per second. Lindy runs at a constant speed of 10 feet per second from Christina to Jack, back to Christina, back to Jack, and so forth. What is the total distance, in feet, that Lindy has traveled when the three meet at one place?

A. 105
B. 210
C. 280
D. 300
E. 420

The relative speed of Jack and Christina is 3 + 4 = 7 feet per second.
The distance between them is 210 feet, hence they will meet in (time) = (distance)/(relative speed) = 210/7 = 30 seconds.

For all this time Lindy was running back and forth, so it covered (distance) = (speed)*(time) = 10*30 = 300 feet.

Answer: D.

Similar question to practice: 12-easy-pieces-or-not-126366.html#p1033924

Hope it helps.

I get the question, and I solved it exactly like you did. However, when you think about it a little more, I feel there is a little more to this question. As Jack and Christina walk towards each other, the distance between them is reducing as well, and the question says Lindy is running back and forth between them. Shouldn't the fact that they are waling towards each other and reducing the distance between them affect the total distance Lindy runs back and forth?

(distance) = (speed)*(time).

We know the speed of the dog and we know the time it runs. We need nothing more.

3 m/s + 4 m/s = 7 m/s * rate at which J & C close the gap between them

210 m / (7 m/s) = 30 s * time period it takes for J & C to meet

30 s x 10 m/s = 300 m * distance that Lindy Covers over a 30s time period

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