# From point A to point B, the distance between which is 30 km, a cyclist and a motorcyclist left at the same time.

**From point A to point B, the distance between which is 30 km, a cyclist and a motorcyclist left at the same time. In an hour, a motorcyclist travels 40 km more than a cyclist. Determine the speed of the cyclist if it is known that he arrived at point B one hour later than the motorcyclist.**

Let the speed of the bike be “x” km / h, so it will take time equal to “30 / x” h.

Then the speed of the motorcycle is “x + 40” km / h, and it will consume “30 / (x + 40)” h.

The motorcycle will spend 1 hour less time. Let’s make the equation:

30 / x = 30 / (x + 40) + 1;

30 / x = (30 + (x + 40)) / (x + 40);

30 / x = (x + 70) / (x + 40);

30 (x + 40) = x (x + 70);

30x + 1200 = x² + 70x;

-x² – 40x + 1200 = 0;

D = (-40) ² – 4 * (-1) * 1200 = 6400;

√D = 80;

x = (40 ± 80) / (2 * (-1));

x₁ = 20 (km / h) bike speed.

x₂ = -60 (does not fit).