One student finished half of the course at 3m / s, and immediately finished the other half at 5m / s______ m/s.
Suppose the total distance is LM, then the first half and the second half of the distance are l2m, the time of the first half of the distance: T1 = SV = 12lm3m / S = 16ls, the time of the second half of the distance: T2 = SV = 12lm5m / S = 110ls, the whole time of the object: T = T1 + T2, the average speed of the whole distance: v = st = st1 + T2 = lm16ls + 110ls = 3.75m/s; so the answer is: 3.75;
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