(2) Point B is when two cars meet. At this time, the local driving time is 4 hours.

(3) Full-time use of the local train 12 hours. The express train reaches point B at point C, the local train continues to point A ... point D, and the local train reaches point A, so the distance between the two cars becomes 900km.

So the speed of the local train is 900/12 = 75 km/h.

When the two cars meet, they have been running for 4 hours at the same time, so the sum of the express speed and the local speed is: V fast +V slow =900/4=225km/h, so the express speed is 225-75 =150 km/h.

(4) When the two cars meet, the distance of the express train is 150km/h*4h=600km, and there is still 900-600=300km from the second place. The express completed the last distance in 300/ 150=2 hours, so the local train also took 2 hours, that is, X = 4+2 = the distance between the two cars at this time is 2*( 150+75)=450km, that is, the coordinate of point C is (6450).

The linear equation passing through b (4 4,0) and c (6 6,450) is y=45x- 180(4≤x≤6).

(5) When the local train meets the second express train, the local train runs for 4.5 hours. Suppose the second express train is x hours later than the first express train, then

900-75x=(4.5-x)*(75+ 150)

solve

X=0.75 hours =45 minutes

Explain the meaning of the equation:

75x is the distance traveled by the local train before the second express train. Subtract from the total distance of 900, and the remaining distance is the distance traveled by the local train and the second express train in the opposite direction. (4.5-x) is the time when the second express train travels in the opposite direction, and (75+ 150) is the time when the second express train travels in the opposite direction.

Answer over!