When a car runs on a straight road, its speed is V1 in the first half of the time and V2 in the second half Find the average speed of the car in this period

When a car runs on a straight road, its speed is V1 in the first half of the time and V2 in the second half Find the average speed of the car in this period

You know the average speed is equal to the total distance divided by the total time, which is v = s / T
If the total time is t, the total distance is equal to the distance taken in the first half of the time plus the distance taken in the second half of the time, that is s = S1 + S2
And S1 = v1t / 2, S2 = v2t / 2
It can be found by substituting the first formula

A car runs 2 / 3 of V1 and 1 / 3 of V2 along the straight road, V2 = 20. If the average speed of the whole journey is 28, find the size of V1

Let the total distance be s,
The first two thirds of the journey time is T1 = (2 / 3S) / v1,
The remaining 1 / 3 journey time is T2 = (1 / 3S) / v2
The average speed of the whole journey v = s / (T1 + T2),
Substituting the data, 28km / h = s / (2 / 3S) / V1 + (1 / 3S) / 20km / h;
So V1 = 35km / h

Let the average speed from a to B be V1, and the average speed from B to a be v2. The average speed of the whole journey is calculated. Suppose the distance is s, T1 = s / V1, T2 = s / V2, and the average speed of the whole journey is 2S / (T1 + T2) = 2S / (s / V1 + S / V2), ← what I don't understand is that 2v1 + 2v2 can be obtained by dividing 2S in this step, but the correct answer is 2v1 * V2 / (V1 + V2), Is there something wrong with me?

If you divide 2S in, you have to
1 / [1 / (2v1) + 1 / (2v2)] = 1 / [(2v1 + 2v2) / 4v1 · V2] = 2v1 · V2 / (V1 + V2)