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

• Last Updated : 22 Jun, 2022

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 = 3
Output: 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 = 3
Input: U = 5, V = 2, X = 4
Output: 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 ` `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

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## Javascript

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Output:

`0.5`

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

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