Bee flying between two trains - upgraded (calculate the speed of the bee)
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So this is a harder version of the classical problem, and I can't figure out how to get started solving it. :(
Two trains 80 kilometers apart are traveling toward each other along the same track. Both trains are going 25 kilometers per hour.
A bee is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until the trains get 100 meters to each other.
The bee completes 36 rounds so it lands 18 times on each train before they are 100 meters from each other.
What is the average speed of the bee (assuming the bee is point-like, no acceleration/decceleration and no extra time when turning)?
Could you help me with this problem as I dont even know how to begin at first place...:(
Thank you in advance!
limits
asked Nov 26 at 16:44
hunsnowboarder
51
add a comment |
up vote
0
down vote
favorite
So this is a harder version of the classical problem, and I can't figure out how to get started solving it. :(
Two trains 80 kilometers apart are traveling toward each other along the same track. Both trains are going 25 kilometers per hour.
A bee is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until the trains get 100 meters to each other.
The bee completes 36 rounds so it lands 18 times on each train before they are 100 meters from each other.
What is the average speed of the bee (assuming the bee is point-like, no acceleration/decceleration and no extra time when turning)?
Could you help me with this problem as I dont even know how to begin at first place...:(
Thank you in advance!
limits
asked Nov 26 at 16:44
hunsnowboarder
51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So this is a harder version of the classical problem, and I can't figure out how to get started solving it. :(
Two trains 80 kilometers apart are traveling toward each other along the same track. Both trains are going 25 kilometers per hour.
A bee is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until the trains get 100 meters to each other.
The bee completes 36 rounds so it lands 18 times on each train before they are 100 meters from each other.
What is the average speed of the bee (assuming the bee is point-like, no acceleration/decceleration and no extra time when turning)?
Could you help me with this problem as I dont even know how to begin at first place...:(
Thank you in advance!
limits
asked Nov 26 at 16:44
hunsnowboarder
51
So this is a harder version of the classical problem, and I can't figure out how to get started solving it. :(
Two trains 80 kilometers apart are traveling toward each other along the same track. Both trains are going 25 kilometers per hour.
A bee is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until the trains get 100 meters to each other.
The bee completes 36 rounds so it lands 18 times on each train before they are 100 meters from each other.
What is the average speed of the bee (assuming the bee is point-like, no acceleration/decceleration and no extra time when turning)?
Could you help me with this problem as I dont even know how to begin at first place...:(
Thank you in advance!
limits
limits
asked Nov 26 at 16:44
hunsnowboarder
51
asked Nov 26 at 16:44
hunsnowboarder
51
asked Nov 26 at 16:44
hunsnowboarder
51
asked Nov 26 at 16:44
hunsnowboarder
51
asked Nov 26 at 16:44
hunsnowboarder
51
51
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The variations of this problem were posted and answered here quite a lot of times.
The first inspiration for you is the graph of position vs time for trains (blue and orange) and the bee (red):
From this image you can geometrically show that all the triangles are similar ($triangle A_1B_1A_2simtriangle A_2B_2A_3$), with ratio between adjacent triangles being the same. So if it takes $t$ minutes for the first round and $qt$ minutes for the second round, it would take $q^2t$ minutes for the 3rd round.
Second inspiration for you is the fact that, moreover, the “tails” of triangle series a similar, i.e. $triangle A_1B_1Osim A_2B_2O$ with the same ratio $q$.
Thus, removing the first round from time interval between start and the meeting of the trains is the same as multiplying it by $q$.
When you remove 18 rounds, it's the same as multiplying by $q^{18}$. How quicker are the trains going to meet when it's just 100 meters between them instead of 80 kilometers? Can you find $q$ from this information?
The final inspiration is the formula of infinite sum of geometric series:
$$
T = t+qt+q^2t+ldots = frac{t}{1-q}
$$
If you know $q$ and the time $T$ until trains meet, can you find the time $t$ between the first and second landing on the first train? If you know the time $t$ can you find the velocity of the bee?
answered Nov 26 at 18:13
Vasily Mitch
1,14827
Thank you so @Vasily Mitch much for your explanation, I fully understand the way I should be thinking about this problem. Thank you so much. However, I am not sure how could I calculate q. So the train when are 80 kms apart and they both go 25 km/h, they would meet in 1.6 hours. If ther is only 0.1 km between them, they would meet in 0.002 hours. Does this mean that $q = frac{0.002}{1.6} = 0.00125$?
– hunsnowboarder
Nov 27 at 14:37
And the other part I don't really understand how to calculate T? Because the formula of infinite sum of geometric series stands for infinite so basically T should be the time till the two trains meet and not when they are 100 meters from each other. Am I right? Thank you again for your tremendous help and clear, understandable explanation!
– hunsnowboarder
Nov 27 at 14:37
It would be if they would be 100m apart only after one round. But it happens after 18 rounds, so $q^{18}=0.1/80$. Yes, you are right about infininity point: the sum calculates the time until they meet.
– Vasily Mitch
Nov 27 at 15:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The variations of this problem were posted and answered here quite a lot of times.
The first inspiration for you is the graph of position vs time for trains (blue and orange) and the bee (red):
From this image you can geometrically show that all the triangles are similar ($triangle A_1B_1A_2simtriangle A_2B_2A_3$), with ratio between adjacent triangles being the same. So if it takes $t$ minutes for the first round and $qt$ minutes for the second round, it would take $q^2t$ minutes for the 3rd round.
Second inspiration for you is the fact that, moreover, the “tails” of triangle series a similar, i.e. $triangle A_1B_1Osim A_2B_2O$ with the same ratio $q$.
Thus, removing the first round from time interval between start and the meeting of the trains is the same as multiplying it by $q$.
When you remove 18 rounds, it's the same as multiplying by $q^{18}$. How quicker are the trains going to meet when it's just 100 meters between them instead of 80 kilometers? Can you find $q$ from this information?
The final inspiration is the formula of infinite sum of geometric series:
$$
T = t+qt+q^2t+ldots = frac{t}{1-q}
$$
If you know $q$ and the time $T$ until trains meet, can you find the time $t$ between the first and second landing on the first train? If you know the time $t$ can you find the velocity of the bee?
answered Nov 26 at 18:13
Vasily Mitch
1,14827
Thank you so @Vasily Mitch much for your explanation, I fully understand the way I should be thinking about this problem. Thank you so much. However, I am not sure how could I calculate q. So the train when are 80 kms apart and they both go 25 km/h, they would meet in 1.6 hours. If ther is only 0.1 km between them, they would meet in 0.002 hours. Does this mean that $q = frac{0.002}{1.6} = 0.00125$?
– hunsnowboarder
Nov 27 at 14:37
And the other part I don't really understand how to calculate T? Because the formula of infinite sum of geometric series stands for infinite so basically T should be the time till the two trains meet and not when they are 100 meters from each other. Am I right? Thank you again for your tremendous help and clear, understandable explanation!
– hunsnowboarder
Nov 27 at 14:37
It would be if they would be 100m apart only after one round. But it happens after 18 rounds, so $q^{18}=0.1/80$. Yes, you are right about infininity point: the sum calculates the time until they meet.
– Vasily Mitch
Nov 27 at 15:21
add a comment |
up vote
0
down vote
accepted
The variations of this problem were posted and answered here quite a lot of times.
The first inspiration for you is the graph of position vs time for trains (blue and orange) and the bee (red):
From this image you can geometrically show that all the triangles are similar ($triangle A_1B_1A_2simtriangle A_2B_2A_3$), with ratio between adjacent triangles being the same. So if it takes $t$ minutes for the first round and $qt$ minutes for the second round, it would take $q^2t$ minutes for the 3rd round.
Second inspiration for you is the fact that, moreover, the “tails” of triangle series a similar, i.e. $triangle A_1B_1Osim A_2B_2O$ with the same ratio $q$.
Thus, removing the first round from time interval between start and the meeting of the trains is the same as multiplying it by $q$.
When you remove 18 rounds, it's the same as multiplying by $q^{18}$. How quicker are the trains going to meet when it's just 100 meters between them instead of 80 kilometers? Can you find $q$ from this information?
The final inspiration is the formula of infinite sum of geometric series:
$$
T = t+qt+q^2t+ldots = frac{t}{1-q}
$$
If you know $q$ and the time $T$ until trains meet, can you find the time $t$ between the first and second landing on the first train? If you know the time $t$ can you find the velocity of the bee?
answered Nov 26 at 18:13
Vasily Mitch
1,14827
Thank you so @Vasily Mitch much for your explanation, I fully understand the way I should be thinking about this problem. Thank you so much. However, I am not sure how could I calculate q. So the train when are 80 kms apart and they both go 25 km/h, they would meet in 1.6 hours. If ther is only 0.1 km between them, they would meet in 0.002 hours. Does this mean that $q = frac{0.002}{1.6} = 0.00125$?
– hunsnowboarder
Nov 27 at 14:37
And the other part I don't really understand how to calculate T? Because the formula of infinite sum of geometric series stands for infinite so basically T should be the time till the two trains meet and not when they are 100 meters from each other. Am I right? Thank you again for your tremendous help and clear, understandable explanation!
– hunsnowboarder
Nov 27 at 14:37
It would be if they would be 100m apart only after one round. But it happens after 18 rounds, so $q^{18}=0.1/80$. Yes, you are right about infininity point: the sum calculates the time until they meet.
– Vasily Mitch
Nov 27 at 15:21
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The variations of this problem were posted and answered here quite a lot of times.
The first inspiration for you is the graph of position vs time for trains (blue and orange) and the bee (red):
From this image you can geometrically show that all the triangles are similar ($triangle A_1B_1A_2simtriangle A_2B_2A_3$), with ratio between adjacent triangles being the same. So if it takes $t$ minutes for the first round and $qt$ minutes for the second round, it would take $q^2t$ minutes for the 3rd round.
Second inspiration for you is the fact that, moreover, the “tails” of triangle series a similar, i.e. $triangle A_1B_1Osim A_2B_2O$ with the same ratio $q$.
Thus, removing the first round from time interval between start and the meeting of the trains is the same as multiplying it by $q$.
When you remove 18 rounds, it's the same as multiplying by $q^{18}$. How quicker are the trains going to meet when it's just 100 meters between them instead of 80 kilometers? Can you find $q$ from this information?
The final inspiration is the formula of infinite sum of geometric series:
$$
T = t+qt+q^2t+ldots = frac{t}{1-q}
$$
If you know $q$ and the time $T$ until trains meet, can you find the time $t$ between the first and second landing on the first train? If you know the time $t$ can you find the velocity of the bee?
answered Nov 26 at 18:13
Vasily Mitch
1,14827
The variations of this problem were posted and answered here quite a lot of times.
The first inspiration for you is the graph of position vs time for trains (blue and orange) and the bee (red):
From this image you can geometrically show that all the triangles are similar ($triangle A_1B_1A_2simtriangle A_2B_2A_3$), with ratio between adjacent triangles being the same. So if it takes $t$ minutes for the first round and $qt$ minutes for the second round, it would take $q^2t$ minutes for the 3rd round.
Second inspiration for you is the fact that, moreover, the “tails” of triangle series a similar, i.e. $triangle A_1B_1Osim A_2B_2O$ with the same ratio $q$.
Thus, removing the first round from time interval between start and the meeting of the trains is the same as multiplying it by $q$.
When you remove 18 rounds, it's the same as multiplying by $q^{18}$. How quicker are the trains going to meet when it's just 100 meters between them instead of 80 kilometers? Can you find $q$ from this information?
The final inspiration is the formula of infinite sum of geometric series:
$$
T = t+qt+q^2t+ldots = frac{t}{1-q}
$$
If you know $q$ and the time $T$ until trains meet, can you find the time $t$ between the first and second landing on the first train? If you know the time $t$ can you find the velocity of the bee?
answered Nov 26 at 18:13
Vasily Mitch
1,14827
answered Nov 26 at 18:13
Vasily Mitch
1,14827
answered Nov 26 at 18:13
Vasily Mitch
1,14827
answered Nov 26 at 18:13
Vasily Mitch
1,14827
1,14827
Thank you so @Vasily Mitch much for your explanation, I fully understand the way I should be thinking about this problem. Thank you so much. However, I am not sure how could I calculate q. So the train when are 80 kms apart and they both go 25 km/h, they would meet in 1.6 hours. If ther is only 0.1 km between them, they would meet in 0.002 hours. Does this mean that $q = frac{0.002}{1.6} = 0.00125$?
– hunsnowboarder
Nov 27 at 14:37
And the other part I don't really understand how to calculate T? Because the formula of infinite sum of geometric series stands for infinite so basically T should be the time till the two trains meet and not when they are 100 meters from each other. Am I right? Thank you again for your tremendous help and clear, understandable explanation!
– hunsnowboarder
Nov 27 at 14:37
It would be if they would be 100m apart only after one round. But it happens after 18 rounds, so $q^{18}=0.1/80$. Yes, you are right about infininity point: the sum calculates the time until they meet.
– Vasily Mitch
Nov 27 at 15:21
add a comment |
Thank you so @Vasily Mitch much for your explanation, I fully understand the way I should be thinking about this problem. Thank you so much. However, I am not sure how could I calculate q. So the train when are 80 kms apart and they both go 25 km/h, they would meet in 1.6 hours. If ther is only 0.1 km between them, they would meet in 0.002 hours. Does this mean that $q = frac{0.002}{1.6} = 0.00125$?
– hunsnowboarder
Nov 27 at 14:37
And the other part I don't really understand how to calculate T? Because the formula of infinite sum of geometric series stands for infinite so basically T should be the time till the two trains meet and not when they are 100 meters from each other. Am I right? Thank you again for your tremendous help and clear, understandable explanation!
– hunsnowboarder
Nov 27 at 14:37
It would be if they would be 100m apart only after one round. But it happens after 18 rounds, so $q^{18}=0.1/80$. Yes, you are right about infininity point: the sum calculates the time until they meet.
– Vasily Mitch
Nov 27 at 15:21
Thank you so @Vasily Mitch much for your explanation, I fully understand the way I should be thinking about this problem. Thank you so much. However, I am not sure how could I calculate q. So the train when are 80 kms apart and they both go 25 km/h, they would meet in 1.6 hours. If ther is only 0.1 km between them, they would meet in 0.002 hours. Does this mean that $q = frac{0.002}{1.6} = 0.00125$?
– hunsnowboarder
Nov 27 at 14:37
Thank you so @Vasily Mitch much for your explanation, I fully understand the way I should be thinking about this problem. Thank you so much. However, I am not sure how could I calculate q. So the train when are 80 kms apart and they both go 25 km/h, they would meet in 1.6 hours. If ther is only 0.1 km between them, they would meet in 0.002 hours. Does this mean that $q = frac{0.002}{1.6} = 0.00125$?
– hunsnowboarder
Nov 27 at 14:37
And the other part I don't really understand how to calculate T? Because the formula of infinite sum of geometric series stands for infinite so basically T should be the time till the two trains meet and not when they are 100 meters from each other. Am I right? Thank you again for your tremendous help and clear, understandable explanation!
– hunsnowboarder
Nov 27 at 14:37
And the other part I don't really understand how to calculate T? Because the formula of infinite sum of geometric series stands for infinite so basically T should be the time till the two trains meet and not when they are 100 meters from each other. Am I right? Thank you again for your tremendous help and clear, understandable explanation!
– hunsnowboarder
Nov 27 at 14:37
It would be if they would be 100m apart only after one round. But it happens after 18 rounds, so $q^{18}=0.1/80$. Yes, you are right about infininity point: the sum calculates the time until they meet.
– Vasily Mitch
Nov 27 at 15:21
It would be if they would be 100m apart only after one round. But it happens after 18 rounds, so $q^{18}=0.1/80$. Yes, you are right about infininity point: the sum calculates the time until they meet.
– Vasily Mitch
Nov 27 at 15:21
add a comment |
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