Someone estimates the acceleration of the train with a watch. He observes it for 3 minutes and finds that the train is 540 meters ahead. After 3 minutes, he observes it for 1 minute and finds that the train is 360 meters ahead. If the train is moving in a straight line with uniform acceleration within 7 minutes, the acceleration of the train is () A. 0.03m/s2B. 0.01m/s2C. 0.5m/s2D. 0.6m/s2

In the first three minutes, the instantaneous velocity at the middle time V1 = 540180m / S = 3m / s, and the instantaneous velocity at the middle time V2 = 36060m / S = 6m / s in one minute. The time interval between the two times is 300s, so a = V2 − v1t = 6 − 3300 = 0.01m/s2

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1. Someone estimates the acceleration of the train with a watch. First, he observes for 3 minutes and finds that the train is moving 540 meters. Then he observes for 1 minute after 3 minutes and the train is moving 360 meters Its analytical formula is a = (360-180) / (6.5-1.5) = 36 m / min * min = 0.6, where 6.5 and 1.5 come from Someone uses a watch to estimate the acceleration of a train. He first observes for 3 minutes and finds that the train is moving 540 meters. After 3 minutes, he observes for 1 minute and the train is moving 360 meters. If the train moves in a straight line with uniform acceleration within 7 minutes, what is the acceleration of the train? A = (360-180) / (6.5-1.5) = 36 m / min * min = 0.6. I just want to know how to calculate 6.5 and 1.5 in this formula. Please answer.
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