Sum of distance traveled between two runners in summation form
$begingroup$
Problem:
Assume two runners $A$ and $B$ are $2h_0$ meters apart. They start running towards each other at a speed of $a$ meters per second. Assume there is a third runner that runs in between the two runners at a speed of $b$ meters per second, with $b > a$. Whenever the third runner meets $A$ or $B$, he turns around and head towards to other person, never changing his speed. What is the total distance traveled by the third runner?
My solution:
Since the third runner is always in between $A$ and $B$, his running session ends when it $A$ meets $B$, which is in the middle, i.e. at position $h_0$.
Since the third runner's position never changes, it is logically the same to ask question of where does the third runner needs to start for all three of the runners to meet in the middle?
If we treat the middle as position $0$, with $A$ starting at $-h_0$ and $B$ starting at $h_0$, then it quickly becomes apparent that the third runner needs to start at a position $frac{bh_0}{a}$ from the middle. Indeed that's the answer.
Question:
How do I represent this in summation form without changing the question of the problem? What about if $A$ and $B$ are running at varying non-linear speed? In other words, how do I represent a piece-wise linear function that's going back and forth and converging between two intersecting monotonic functions?
summation
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
$endgroup$
add a comment |
$begingroup$
Problem:
Assume two runners $A$ and $B$ are $2h_0$ meters apart. They start running towards each other at a speed of $a$ meters per second. Assume there is a third runner that runs in between the two runners at a speed of $b$ meters per second, with $b > a$. Whenever the third runner meets $A$ or $B$, he turns around and head towards to other person, never changing his speed. What is the total distance traveled by the third runner?
My solution:
Since the third runner is always in between $A$ and $B$, his running session ends when it $A$ meets $B$, which is in the middle, i.e. at position $h_0$.
Since the third runner's position never changes, it is logically the same to ask question of where does the third runner needs to start for all three of the runners to meet in the middle?
If we treat the middle as position $0$, with $A$ starting at $-h_0$ and $B$ starting at $h_0$, then it quickly becomes apparent that the third runner needs to start at a position $frac{bh_0}{a}$ from the middle. Indeed that's the answer.
Question:
How do I represent this in summation form without changing the question of the problem? What about if $A$ and $B$ are running at varying non-linear speed? In other words, how do I represent a piece-wise linear function that's going back and forth and converging between two intersecting monotonic functions?
summation
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
$endgroup$
$begingroup$
It doesn't matter where the third runner starts, they will always meet in the middle given A and B are equal distances from the middle and run at the same speed. The 3rd runner never moves from being in between A and B. All three runners run for the same amount of time. Is a the closing speed or the speed of each runner A and B? Assuming a is the closing speed, the time it takes for A and B to meet is $t = frac{2h_0}{a}$ secs. The distance the third runner travels is therefore $s = bcdot frac{2h_0}{a}$ m. If a is the individual speeds of A and B, then it's $s = bcdot frac{2h_0}{2a}$ m
$endgroup$
– Phil H
Dec 22 '18 at 1:09
$begingroup$
This is similar to the bee-flying-between-two-trains problem.
$endgroup$
– hypergeometric
Jan 21 at 13:44
add a comment |
$begingroup$
Problem:
Assume two runners $A$ and $B$ are $2h_0$ meters apart. They start running towards each other at a speed of $a$ meters per second. Assume there is a third runner that runs in between the two runners at a speed of $b$ meters per second, with $b > a$. Whenever the third runner meets $A$ or $B$, he turns around and head towards to other person, never changing his speed. What is the total distance traveled by the third runner?
My solution:
Since the third runner is always in between $A$ and $B$, his running session ends when it $A$ meets $B$, which is in the middle, i.e. at position $h_0$.
Since the third runner's position never changes, it is logically the same to ask question of where does the third runner needs to start for all three of the runners to meet in the middle?
If we treat the middle as position $0$, with $A$ starting at $-h_0$ and $B$ starting at $h_0$, then it quickly becomes apparent that the third runner needs to start at a position $frac{bh_0}{a}$ from the middle. Indeed that's the answer.
Question:
How do I represent this in summation form without changing the question of the problem? What about if $A$ and $B$ are running at varying non-linear speed? In other words, how do I represent a piece-wise linear function that's going back and forth and converging between two intersecting monotonic functions?
summation
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
$endgroup$
Problem:
Assume two runners $A$ and $B$ are $2h_0$ meters apart. They start running towards each other at a speed of $a$ meters per second. Assume there is a third runner that runs in between the two runners at a speed of $b$ meters per second, with $b > a$. Whenever the third runner meets $A$ or $B$, he turns around and head towards to other person, never changing his speed. What is the total distance traveled by the third runner?
My solution:
Since the third runner is always in between $A$ and $B$, his running session ends when it $A$ meets $B$, which is in the middle, i.e. at position $h_0$.
Since the third runner's position never changes, it is logically the same to ask question of where does the third runner needs to start for all three of the runners to meet in the middle?
If we treat the middle as position $0$, with $A$ starting at $-h_0$ and $B$ starting at $h_0$, then it quickly becomes apparent that the third runner needs to start at a position $frac{bh_0}{a}$ from the middle. Indeed that's the answer.
Question:
How do I represent this in summation form without changing the question of the problem? What about if $A$ and $B$ are running at varying non-linear speed? In other words, how do I represent a piece-wise linear function that's going back and forth and converging between two intersecting monotonic functions?
summation
summation
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
edited Dec 22 '18 at 0:08
B.Li
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
asked Dec 22 '18 at 0:03
B.LiB.Li
413210
413210
$begingroup$
It doesn't matter where the third runner starts, they will always meet in the middle given A and B are equal distances from the middle and run at the same speed. The 3rd runner never moves from being in between A and B. All three runners run for the same amount of time. Is a the closing speed or the speed of each runner A and B? Assuming a is the closing speed, the time it takes for A and B to meet is $t = frac{2h_0}{a}$ secs. The distance the third runner travels is therefore $s = bcdot frac{2h_0}{a}$ m. If a is the individual speeds of A and B, then it's $s = bcdot frac{2h_0}{2a}$ m
$endgroup$
– Phil H
Dec 22 '18 at 1:09
$begingroup$
This is similar to the bee-flying-between-two-trains problem.
$endgroup$
– hypergeometric
Jan 21 at 13:44
add a comment |
$begingroup$
It doesn't matter where the third runner starts, they will always meet in the middle given A and B are equal distances from the middle and run at the same speed. The 3rd runner never moves from being in between A and B. All three runners run for the same amount of time. Is a the closing speed or the speed of each runner A and B? Assuming a is the closing speed, the time it takes for A and B to meet is $t = frac{2h_0}{a}$ secs. The distance the third runner travels is therefore $s = bcdot frac{2h_0}{a}$ m. If a is the individual speeds of A and B, then it's $s = bcdot frac{2h_0}{2a}$ m
$endgroup$
– Phil H
Dec 22 '18 at 1:09
$begingroup$
This is similar to the bee-flying-between-two-trains problem.
$endgroup$
– hypergeometric
Jan 21 at 13:44
$begingroup$
It doesn't matter where the third runner starts, they will always meet in the middle given A and B are equal distances from the middle and run at the same speed. The 3rd runner never moves from being in between A and B. All three runners run for the same amount of time. Is a the closing speed or the speed of each runner A and B? Assuming a is the closing speed, the time it takes for A and B to meet is $t = frac{2h_0}{a}$ secs. The distance the third runner travels is therefore $s = bcdot frac{2h_0}{a}$ m. If a is the individual speeds of A and B, then it's $s = bcdot frac{2h_0}{2a}$ m
$endgroup$
– Phil H
Dec 22 '18 at 1:09
$begingroup$
It doesn't matter where the third runner starts, they will always meet in the middle given A and B are equal distances from the middle and run at the same speed. The 3rd runner never moves from being in between A and B. All three runners run for the same amount of time. Is a the closing speed or the speed of each runner A and B? Assuming a is the closing speed, the time it takes for A and B to meet is $t = frac{2h_0}{a}$ secs. The distance the third runner travels is therefore $s = bcdot frac{2h_0}{a}$ m. If a is the individual speeds of A and B, then it's $s = bcdot frac{2h_0}{2a}$ m
$endgroup$
– Phil H
Dec 22 '18 at 1:09
$begingroup$
This is similar to the bee-flying-between-two-trains problem.
$endgroup$
– hypergeometric
Jan 21 at 13:44
$begingroup$
This is similar to the bee-flying-between-two-trains problem.
$endgroup$
– hypergeometric
Jan 21 at 13:44
add a comment |
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$begingroup$
It doesn't matter where the third runner starts, they will always meet in the middle given A and B are equal distances from the middle and run at the same speed. The 3rd runner never moves from being in between A and B. All three runners run for the same amount of time. Is a the closing speed or the speed of each runner A and B? Assuming a is the closing speed, the time it takes for A and B to meet is $t = frac{2h_0}{a}$ secs. The distance the third runner travels is therefore $s = bcdot frac{2h_0}{a}$ m. If a is the individual speeds of A and B, then it's $s = bcdot frac{2h_0}{2a}$ m
$endgroup$
– Phil H
Dec 22 '18 at 1:09
$begingroup$
This is similar to the bee-flying-between-two-trains problem.
$endgroup$
– hypergeometric
Jan 21 at 13:44