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<meta property="og:description" content="Athletes run in a column of length l at speed v. The coach runs towards them with speed u < v. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?">
<title>Athletes run in a column of length l at speed v. The coach runs towards them with speed u < v. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?</title>
$1.1.6.$ Athletes run in a column of length $l$ at speed $v$. The coach runs towards them with speed $u < v$. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?
</p>
<h3>Solution</h3>
<p>
<p class="TxtSolutions"> Let us imagine stopping the column, then the coach besides his speed will have the speed of the column directed in the opposite direction. With this relative velocity $v + u$ he will during time
</p>
<p style="text-align: center;">
$$t = \frac{l}{v + u}$$
</p>
<p class="TxtSolutions"> will run along the column and equalise with the tail. The head of the column, having turned round, will move with relative velocity $v - u$ and after time
</p>
<p style="text-align: center;">
$$t = \frac{{l}'}{v - u}$$
</p>
<p class="TxtSolutions"> where ${l}'$ is the length of the new “column” after overtaking. Then
</p>
<p style="text-align: center;">
$$\frac{{l}'}{v - u} = \frac{l}{v + u}$$
</p>
<p class="TxtSolutions"> and the length of the new column
</p>
<p style="text-align: center;">
$$\fbox{${l}' = \frac{v - u}{v + u}l$}$$
</p>
</p>
<h4>Answer</h4>
<p>
When all athletes turn around, the length of the new column will be equal to ${l}' = \frac{v - u}{v + u}l$
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
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<meta name="description" content="The largest dataset of solutions of 'Savchenko. Problems in Physics'. Savchenko’s Problems in General Physics is widely used to prepare for olympiads and it is a useful tool to
<meta name="description" content="Athletes run in a column of length l at speed v. The coach runs towards them with speed u < v. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?">
master and sharpen your skills and techniques in comptetitive problem solving. Some of these problems were a source
of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
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<meta property="og:description" content="Athletes run in a column of length l at speed v. The coach runs towards them with speed u < v. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?">
<title>Athletes run in a column of length l at speed v. The coach runs towards them with speed u < v. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?</title>
$1.1.6.$ Athletes run in a column of length $l$ at speed $v$. The coach runs towards them with speed $u < v$. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?
$1.1.6.$ Athletes run in a column of length $l$ at speed $v$. The coach runs towards them with speed $u < v$. Each athlete, when he reaches the coach, turns around and starts running back with the same speed. What is the length of the column when all the athletes turn around?
</p>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
<p class="TxtSolutions"> Let us imagine stopping the column, then the coach besides his speed will have the speed of the column directed in the opposite direction. With this relative velocity $v + u$ he will during time
<p class="TxtSolutions"> Let us imagine stopping the column, then the coach besides his speed will have the speed of the column directed in the opposite direction. With this relative velocity $v + u$ he will during time
</p>
</p>
<p style="text-align: center;">
<p style="text-align: center;">
$$t = \frac{l}{v + u}$$
$$t = \frac{l}{v + u}$$
</p>
</p>
<p class="TxtSolutions"> will run along the column and equalise with the tail. The head of the column, having turned round, will move with relative velocity $v - u$ and after time
<p class="TxtSolutions"> will run along the column and equalise with the tail. The head of the column, having turned round, will move with relative velocity $v - u$ and after time
</p>
</p>
<p style="text-align: center;">
<p style="text-align: center;">
$$t = \frac{{l}'}{v - u}$$
$$t = \frac{{l}'}{v - u}$$
</p>
</p>
<p class="TxtSolutions"> where ${l}'$ is the length of the new “column” after overtaking. Then
<p class="TxtSolutions"> where ${l}'$ is the length of the new “column” after overtaking. Then
</p>
</p>
<p style="text-align: center;">
<p style="text-align: center;">
$$\frac{{l}'}{v - u} = \frac{l}{v + u}$$
$$\frac{{l}'}{v - u} = \frac{l}{v + u}$$
</p>
</p>
<p class="TxtSolutions"> and the length of the new column
<p class="TxtSolutions"> and the length of the new column
</p>
</p>
<p style="text-align: center;">
<p style="text-align: center;">
$$\fbox{${l}' = \frac{v - u}{v + u}l$}$$
$$\fbox{${l}' = \frac{v - u}{v + u}l$}$$
</p>
</p>
</p>
</p>
<h4>Answer</h4>
<h4>Answer</h4>
<p>
<p>
When all athletes turn around, the length of the new column will be equal to ${l}' = \frac{v - u}{v + u}l$
When all athletes turn around, the length of the new column will be equal to ${l}' = \frac{v - u}{v + u}l$
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>