David walks his way to library every day. Generally, his speed is about 2.5 m/s but if he walks at speed of 3.5 m/s he will reach the library 5 minutes early. How far is the library from his home?

To solve this problem, we need to understand that the distance to the library remains the same regardless of how fast David walks. We can use the formula for distance which is speed multiplied by time.

1. First, we need to convert the time difference from minutes to seconds because the speed is given in meters per second. 5 minutes is equal to 300 seconds.

2. We know that the time it takes David to walk to the library at his usual speed (2.5 m/s) is 300 seconds more than the time it takes him to walk at the faster speed (3.5 m/s). We can express this as an equation: t + 300 = t1, where t is the time it takes David to walk at the faster speed and t1 is the time it takes him to walk at his usual speed.

3. We also know that the distance to the library is the same regardless of how fast David walks. We can express this as another equation: 2.5 * t1 = 3.5 * t.

4. Now we can solve these two equations simultaneously. Substituting the first equation into the second gives us: 2.5 * (t + 300) = 3.5 * t.

5. Simplifying this equation gives us: 2.5t + 750 = 3.5t.

6. Solving for t gives us: t = 750 seconds.

7. Now we can find the distance to the library by substituting t back into the equation for distance: d = 3.5 * t = 3.5 * 750 = 2625 meters.

So, the library is 2625 meters away from David's home.