A convoy of cars 900 m long is moving along the road at a speed of 12 m / s. A motorcyclist is sent from the lead

A convoy of cars 900 m long is moving along the road at a speed of 12 m / s. A motorcyclist is sent from the lead car with an order to the last car and immediately comes back. The speed of the motorcyclist relative to the ground is 13 m / s. How long will a motorcyclist spend on such a movement?

S = 900 m.

Vk = 12 m / s.

Vm = 13 m / s.

t -?

The travel time of the motorcycle t will be the sum of the travel time from the first car to the last t1 and the travel time from the last to the first t2: t = t1 + t2.

During time t1, the motorcyclist covers a path equal to the length of the column S.

Since it moves in the opposite direction of the column movement, its speed of movement relative to the column Vm1 will be: Vm1 = Vk + Vm.

t1 = S / (Vk + Vm).

t1 = 900 m / (12 m / s + 13 m / s) = 36 s.

During the time t2, the motorcyclist covers a path equal to the length of the column S.

Since it moves in the direction of the column movement, its speed of movement relative to the column Vm2 will be: Vm2 = Vm – Vk.

t2 = S / (Vm – Vk).

t2 = 900 m / (13 m / s – 12 m / s) = 900 s.

t = 36 s + 900 s = 936 s.

Answer: the motorcyclist will spend time t = 936 s on the movement.