The distance between cities A and B is 70 km. A car left town A for town B, and 15 minutes later a motorcyclist

The distance between cities A and B is 70 km. A car left town A for town B, and 15 minutes later a motorcyclist followed it at a speed of 50 km / h. The motorcyclist caught up with the car in city C and turned back. When he was halfway from C to A, the car arrived at B. Find the distance from A to C.

Let’s denote the time during which the motorcycle travels the distance from A to C as t.

Then the distance AC = (50 x t) km.

The car travels the distance from A to C in (t + 0.25) hours, and the distance from C to B in (t / 2) hours.

Then the distance from A to B cars travels for (t + 0.25 + 0.5 x t) = ((1.5 x t) + 0.25).

Then the vehicle speed will be equal to: Va = 70 / ((1.5 x t) + 0.25).

Knowing the speed of the car and the time during which it traveled the distance from A to C, we express AC.

AC = (70 / ((1.5 x t) + 0.25)) x (t + 0.25).

Then:

(70 / ((1.5 x t) + 0.25)) x (t + 0.25) = 50 x t.

70 x (t + 0.25) = ((1.5 x t) + 0.25) x 50 x t.

70 x t + 17.5 = 75 x t2 + 12.5 x t.

75 x t2 – 57.5 x t – 17.5 = 0.

3 x t2 – 2.3 x t – 0.7 = 0.

Let’s solve the quadratic equation.

D = b2 – 4 x a x c = (-2.3) ^ 2 – 4 x 3 x (-0.7) = 5.29 + 8.4 = 13.69.

t1 = (2.3 – √13.69) / (2 x 3) = (2.3 – 3.7) / 6 = -1.4 / 6 = -0.23. (Doesn’t match because <0).

t2 = (2.3 + √13.69) / (2 x 3) = (2.3 + 3.7) / 6 = 6/6 = 1.0 h.

Then the distance from A to C will be equal to:

AC = 50 x 1 = 50 km.

Answer: The distance from A to C is 50 km.