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
RELATED INFORMATIONS
- 1. Someone uses a watch to measure the acceleration of a train. First, measure it for three minutes. The train runs 540 meters. After three minutes, measure it for another minute. The train runs 360 meters. If the train is seven minutes Someone uses a watch to measure the acceleration of a train. First, he measures it for three minutes. The train runs 540 meters. Then, he measures it for another minute after three minutes. The train runs 360 meters. If the train accelerates in a straight line within seven minutes, can he calculate the acceleration of the train?
- 2. Someone used a watch to estimate the acceleration of the train, first observed for 3 minutes, and found that the train was moving 540 meters; then observed for 1 minute after 3 minutes, and found that the train was moving 360 meters Suppose the train moves in a straight line with uniform acceleration, what is the acceleration of the train estimated by the person
- 3. 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
- 4. 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
- 5. 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.
- 6. A train set out from the station to do uniform acceleration linear motion, 0.5m/s, at this time there is just a A train starts from the station and makes a uniform acceleration linear motion, 0.5m/s. At this time, a bicycle (which can be regarded as a particle) passes by the locomotive, V0 = 8m / s, and the length of the train is L = 336m?
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