# Time until distance gets equal to X between two objects moving in opposite direction

Consider two people moving in opposite direction with speeds **U meters/second** and **V meters/second** respectively. The task is to find how long will take to make the distance between them X meters.**Examples:**

Input:U = 3, V = 3, X = 3Output:0.5

After 0.5 seconds, policeman A will be at distance 1.5 meters

and policeman B will be at distance 1.5 meters in the opposite direction

The distance between the two policemen is 1.5 + 1.5 = 3Input:U = 5, V = 2, X = 4Output:0.571429

**Approach:** It can be solved using **distance = speed * time**. Here, distance would be equal to the given range i.e. **distance = X** and speed would be the sum of the two speeds because they are moving in the opposite direction i.e. **speed = U + V**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the time for which` `// the two policemen can communicate` `double` `getTime(` `int` `u, ` `int` `v, ` `int` `x)` `{` ` ` `double` `speed = u + v;` ` ` `// time = distance / speed` ` ` `double` `time` `= x / speed;` ` ` `return` `time` `;` `}` `// Driver code` `int` `main()` `{` ` ` `double` `u = 3, v = 3, x = 3;` ` ` `cout << getTime(u, v, x);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG ` `{` ` ` `// Function to return the time for which` ` ` `// the two policemen can communicate` ` ` `static` `double` `getTime(` `int` `u, ` `int` `v, ` `int` `x) ` ` ` `{` ` ` `double` `speed = u + v;` ` ` `// time = distance / speed` ` ` `double` `time = x / speed;` ` ` `return` `time;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{` ` ` `int` `u = ` `3` `, v = ` `3` `, x = ` `3` `;` ` ` `System.out.println(getTime(u, v, x));` ` ` `}` `}` `/* This code contributed by PrinciRaj1992 */` |

## Python3

`# Python3 implementation of the approach ` `# Function to return the time ` `# for which the two policemen ` `# can communicate ` `def` `getTime(u, v, x): ` ` ` `speed ` `=` `u ` `+` `v ` ` ` `# time = distance / speed ` ` ` `time ` `=` `x ` `/` `speed ` ` ` `return` `time ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `u, v, x ` `=` `3` `, ` `3` `, ` `3` ` ` `print` `(getTime(u, v, x)) ` `# This code is contributed` `# by Rituraj Jain` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG ` `{` ` ` `// Function to return the time for which` ` ` `// the two policemen can communicate` ` ` `static` `double` `getTime(` `int` `u, ` `int` `v, ` `int` `x) ` ` ` `{` ` ` `double` `speed = u + v;` ` ` `// time = distance / speed` ` ` `double` `time = x / speed;` ` ` `return` `time;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main() ` ` ` `{` ` ` `int` `u = 3, v = 3, x = 3;` ` ` `Console.WriteLine(getTime(u, v, x));` ` ` `}` `}` `// This code is contributed` `// by Akanksha Rai` |

## PHP

`<?php` `// PHP implementation of the approach` `// Function to return the time for which` `// the two policemen can communicate` `function` `getTime(` `$u` `, ` `$v` `, ` `$x` `)` `{` ` ` `$speed` `= ` `$u` `+ ` `$v` `;` ` ` `// time = distance / speed` ` ` `$time` `= ` `$x` `/ ` `$speed` `;` ` ` `return` `$time` `;` `}` `// Driver code` `$u` `= 3; ` `$v` `= 3; ` `$x` `= 3;` `echo` `getTime(` `$u` `, ` `$v` `, ` `$x` `);` `// This code is contributed ` `// by Akanksha Rai` `?>` |

## Javascript

`<script>` `// JavaScript implementation of the approach` `// Function to return the time for which` `// the two policemen can communicate` `function` `getTime(u, v, x)` `{` ` ` `let speed = u + v;` ` ` `// time = distance / speed` ` ` `let time = x / speed;` ` ` `return` `time;` `}` `// Driver code` ` ` `let u = 3, v = 3, x = 3;` ` ` `document.write(getTime(u, v, x));` `// This code is contributed by Surbhi Tyagi.` `</script>` |

**Output:**

0.5

**Time Complexity:** O(1)**Auxiliary Space: **O(1)