Please help with this question

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damndaniel09

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A person X is chasing a thief Y. X is running at 20m/s and Y at 10m/s speed. Y has a head start of 5000m. A dog with a speed of 30m/s is running to-and-fro between X and Y, starting from X. What is the distance in forward direction covered by the dog by the time X catches Y? (Forward direction being the direction in which X and Y are running)

Been stuck on this for a long time. Please help.

Thank you in advance!
 
damndaniel09 said:
A person X is chasing a thief Y. X is running at 20m/s and Y at 10m/s speed. Y has a head start of 5000m. A dog with a speed of 30m/s is running to-and-fro between X and Y, starting from X. What is the distance in forward direction covered by the dog by the time X catches Y? (Forward direction being the direction in which X and Y are running)

Been stuck on this for a long time. Please help.

Thank you in advance!
Click to expand...
First calculate the time required by X to catch-up with Y.

Please share your work/thoughts about this problem with us.
 
I found out the time taken by X to catch-up to Y by using relative velocity and it is 500s.

In this time the total distance travelled by the dog will be 30*500 = 15000m.
If we take forward distance as F and backward distance B, then F+B = 15000.
I think if we can find another equation using F and B then it can be solved. Right now I am stuck here.
 
I found an answer. I'll outline my method below. However, does anyone have a simpler method? I wouldn't want to advise OP to use this if there's a simpler way! Also @damndaniel09 please let us know if you understand/ would be able to use this strategy (and if you think that there ought to be a simpler way). What have you been studying in class recently, and would you expect to use a method like this?

--

My strategy was to consider a single "cycle" as the dog going from X to Y and then from Y back to X (at which time X will be at a different position than they were at the start of the cycle).

1. Find an expression for how long a cycle takes in terms of the cycle's starting time.
2. Find an expression for the forward distance traveled by the dog, during one cycle, in terms of the cycle's starting time.
3. Using 1, find an expression for the staring time of cycle "n" (in terms of n). This involves solving a recursive sequence
4. Using 2 and 3,write a sum for the total forward distance from cycle 0 to cycle "n". Solving this giving an expression in terms of n
5. Using 4, find the limit of the total distance as "n" tends to infinity (this step is easy)

I found the result to be an integer (somewhere below 15,000m :thumbup:)
 
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Cubist said:
I found an answer. I'll outline my method below. However, does anyone have a simpler method? I wouldn't want to advise OP to use this if there's a simpler way! Also @damndaniel09 please let us know if you understand/ would be able to use this strategy (and if you think that there ought to be a simpler way). What have you been studying in class recently, and would you expect to use a method like this?

--

My strategy was to consider a single "cycle" as the dog going from X to Y and then from Y back to X (at which time X will be at a different position than they were at the start of the cycle).

1. Find an expression for how long a cycle takes in terms of the cycle's starting time.
2. Find an expression for the forward distance traveled by the dog, during one cycle, in terms of the cycle's starting time.
3. Using 1, find an expression for the staring time of cycle "n" (in terms of n). This involves solving a recursive sequence
4. Using 2 and 3,write a sum for the total forward distance from cycle 0 to cycle "n". Solving this giving an expression in terms of n
5. Using 4, find the limit of the total distance as "n" tends to infinity (this step is easy)

I found the result to be an integer (somewhere below 15,000m :thumbup:)
Click to expand...
Thank you for the reply!

Actually there should be an easier way to solve this question. This question was intended for a competitive exam and each question is usually given 1 to 2 mins for solving. The solution you have provided looks a bit tedious for that time limit. I haven't tried solving it by that method. Will let you know once I solve it.
 
Cubist said:
I found an answer. I'll outline my method below. However, does anyone have a simpler method? I wouldn't want to advise OP to use this if there's a simpler way! Also @damndaniel09 please let us know if you understand/ would be able to use this strategy (and if you think that there ought to be a simpler way). What have you been studying in class recently, and would you expect to use a method like this?
Click to expand...

If you consider one trip from X to Y and back to X, you can find what fraction of the total time (and hence of the total distance) is spent going forward. That fraction of the total distance will be the answer.

(A time-distance graph can help you see why this ratio is the same for each cycle.)
 
Dr.Peterson said:
If you consider one trip from X to Y and back to X, you can find what fraction of the total time (and hence of the total distance) is spent going forward. That fraction of the total distance will be the answer.

(A time-distance graph can help you see why this ratio is the same for each cycle.)
Click to expand...
Brilliant!

I like this problem. It seems similar to this puzzle that has a paradoxical element https://mathworld.wolfram.com/TwoTrainsPuzzle.html. Can the trains ever meet because the fly will have had to go back and forth an infinite number of times before the trains crash :)
 
Cubist said:
Brilliant!

I like this problem. It seems similar to this puzzle that has a paradoxical element https://mathworld.wolfram.com/TwoTrainsPuzzle.html. Can the trains ever meet because the fly will have had to go back and forth an infinite number of times before the trains crash :)
Click to expand...
Yes, I am familiar with the fly problem, but I don't think I've see this forward-only variant.

My first thought was just to say infinite series aren't that hard, but then this idea popped into my head, I sketched the graph to convince myself it was valid, and then solved it to make sure it wasn't harder than I thought (and gave an integer answer).

To my shame, I haven't tried using the series ...
 
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