A meteorological research center observed the whole process of a sandstorm from its occurrence to its end. At the beginning, the wind speed increased by 2km / h on average. After 4h, the wind speed increased by 4km / h on average when the sandstorm passed through the open desert. For a period of time, the wind speed remained unchanged at 32km / h. When the sandstorm met the green vegetation area, the wind speed decreased by 1km / h on average. At 40H, the measured wind speed was 17km / h According to the images of wind speed and time, the following questions are answered: (1) fill in the corresponding values in the y-axis (); (2) how many hours have the sandstorm passed from the occurrence to the end? (3) The functional relationship between wind speed y (km / h) and time x (H) is obtained when x ≥ 4

A meteorological research center observed the whole process of a sandstorm from its occurrence to its end. At the beginning, the wind speed increased by 2km / h on average. After 4h, the wind speed increased by 4km / h on average when the sandstorm passed through the open desert. For a period of time, the wind speed remained unchanged at 32km / h. When the sandstorm met the green vegetation area, the wind speed decreased by 1km / h on average. At 40H, the measured wind speed was 17km / h According to the images of wind speed and time, the following questions are answered: (1) fill in the corresponding values in the y-axis (); (2) how many hours have the sandstorm passed from the occurrence to the end? (3) The functional relationship between wind speed y (km / h) and time x (H) is obtained when x ≥ 4

(1) The wind speed is 2 × 4 = 8km / h in 4 hours and 8 + 4 × (10-4) = 32km / h in 10 hours. (2) if the deceleration time is x, then 32 + (- 1) · x = 0, and the solution is x = 32H. The sandstorm lasts 25 + 32 = 57h from the occurrence to the end. (3) if the analytic formula is y = KX + B, when 4 ≤ x ≤ 10, substitute (4,8), (10,32) into y = KX + B, and the solution is k = 4B = - 8 When 10 ＜ x ≤ 25, because the wind speed does not change, y = 32, 10 ＜ x ≤ 25; when 25 ＜ x ≤ 57, substituting (25, 32), (57, 0) into y = KX + B, 25K + B = 32 ① 57K + B = 0 ②, the solution is k = - 1B = 57, so the analytical formula is y = - x + 57, 25 ＜ x ≤ 57