The speed of the train is 8 m / s. After the engine is turned off, the speed will be reduced to 6 m / s when the train moves forward for 70 M. if the train goes through another 50 s, the distance of the train is () A. 50mB. 90mC. 120mD. 160m

The speed of the train is 8 m / s. After the engine is turned off, the speed will be reduced to 6 m / s when the train moves forward for 70 M. if the train goes through another 50 s, the distance of the train is () A. 50mB. 90mC. 120mD. 160m

Let the acceleration of the train be a, according to vt2 − & nbsp; V02 & nbsp; = 2aX, we can get a = vt2 − v022x = 36 − 642 × 70 = - 15m / S2, so the time from 6m / s to stop is t = △ VA = − 6 − 15 = 30s. Therefore, after another 50s, the distance of the train is actually 30s, s =. VT = 62 × 30 = 90m