# From point A to point B a motorcyclist drove off, driving 48 km per hour. After 45 minutes, another motorcyclist left

**From point A to point B a motorcyclist drove off, driving 48 km per hour. After 45 minutes, another motorcyclist left B for A at a speed of 50 km / h. Knowing that the distance between A and B is 330 km, find how far from A will motorcyclists meet?**

Let S1 be the distance traveled by the first rider before meeting the second rider, and S2 be the distance traveled by the second. The sum of these distances will be the distance between points A and B:

S1 + S2 = S

The distance of the first rider can be calculated using the formula:

S1 = V1t1,

and the distance of the second:

S2 = V2t2

Since the second motorcyclist left 45 minutes later than the first, his time can be expressed in terms of t1:

t2 = t1 +45 min = t1 + 45/60 = t1 +3/4 h.

Substitute the resulting expressions into the first equation:

48t1 + 50 (t1 + 3/4) = 330;

48t1 + 50t1 + 150/4 = 330;

98t1 + 75/2 = 330;

196t1 + 75 = 660;

196t1 = 585;

t1 = 585/196;

Let’s find now S1, or the distance from A, where the motorcyclists will meet:

S1 = V1t1 = 48 * 585/196 = 28080/196 = 143 52/196 = 143 13/49 km.

Answer: motorcyclists will meet from point A at a distance of 143 13/49 km.