# The convoy of cars is moving along the road at a speed of 40 km / h. The messenger left the end of the column at its beginning

The convoy of cars is moving along the road at a speed of 40 km / h. The messenger left the end of the column at its beginning, passed on the plan for further movement and returned back. He spent 1 hour and 12 minutes for the entire journey. Find the speed of the messenger if the column is 20 km long.

1. Speed of movement of the convoy of cars: Vк = 40 km / h;
2. Messenger speed: VP km / h;
3. column length: L = 20 km;
4. Travel time of the messenger to the head of the column and back: T = 1 hour 12 minutes = (6/5) hour;
T = T1 + T2 = (L / (Vp – Vk)) + (L / (Vp + Vk)) =
(40 * Vp) / (Vn ^ 2 – Vk ^ 2) = 6/5;
200 * Vp = 6 * (Vp ^ 2 – Vk ^ 2);
3 * Vp ^ 2 – 100 * Vp – 4800 = 0;
Vp1,2 = (100 + – sqrt (100 ^ 2 + 4 * 3 * 4800)) / (2 * 3) = (100 + -260) / 6;
A negative root is meaningless;
Vп = (100 +260) / 6 = 69 km / h.
Answer: the speed of the messenger is 60 km / h.

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