When two passengers get off the train, they follow the same route to the same place where the station is s. The speed of passenger a is V1 in the first half of the time and V2 in the other half. Passenger B uses V1 to complete the first half of the journey and V2 to complete the second half. If it takes T1 and T2 for passenger a and passenger B to arrive at the same place from the station, then T1 =? T2 =?

When two passengers get off the train, they follow the same route to the same place where the station is s. The speed of passenger a is V1 in the first half of the time and V2 in the other half. Passenger B uses V1 to complete the first half of the journey and V2 to complete the second half. If it takes T1 and T2 for passenger a and passenger B to arrive at the same place from the station, then T1 =? T2 =?

A: V1 * (1 / 2t1) + V2 * (1 / 2t1) = s, T1 = 2S / (V1 + V2)
B: T2 = 1 / 2S / V1 + 1 / 2S / V2 = s (V1 + V2) / (2v1v2)

Simple motion analysis
If a triangle block a is placed on the rough horizontal plane, and B slides uniformly on the inclined plane of a, how does a move
Who can help me analyze the contradiction between Newton's law and momentum treatment

Let me analyze it
Newton's Law: A, B as a whole, no acceleration. So the external force on the system is 0. So a is not affected by the friction of the ground. That is, a is stationary
Momentum: in the horizontal direction, using the previous conclusion, the system is not subject to external force, so momentum is conserved. The horizontal momentum of B is unchanged, so is A. It is still static, so it is not contradictory

A simple physical motion problem
I'm sorry, but I'm going to raise one. I've done some questions first
When a yacht sails against the water at a constant speed, it loses a life buoy somewhere. It is found T seconds after the loss, so it immediately returns to catch up. At the downstream s point, it is found that the water flow speed is set constant, the yacht's round-trip speed is set constant, and the turning time is ignored

The speed of boat to water is constant, and the speed of circle is the speed of water
The retrograde time is t, and the anterograde time is t. The total time of the circle is 2T. The displacement of the circle is s
Therefore, the velocity of the circle is v = s / (2t) -- -- that is, the velocity of water