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1.以10m/s的速度行驶的汽车,紧急刹车后加速度的大小为2m/s2,求刹车后6.0s末的速度和位移 错解 vt=v0+at=10m/s+(-2)m/s2×6s=-2m/s. s=v0t+1/2at2=10m/s×6s+1/2×(-2)m/s2×(6s)2 =24m. 分析这种解法错在盲目用题给时间6秒去套用公式.本题中汽车从刹车到停止不要6秒时间,只要5秒时间,有1秒钟是停在原地不动的,解这类题目的一般步骤是: (1)根据公式t停=0-v0/a=-v0/a计算出停车所需时间; (2)比较题目所给时间t与停车时间t停的大小; (3)根据t与t停的大小,确定用下面哪种方法进
1. For a car traveling at a speed of 10m/s, the magnitude of the acceleration after an emergency brake is 2m/s2. Calculate the speed and displacement at the end of 6.0s after the brake. vt=v0+at=10m/s+(-2)m/ S2 × 6s = -2m/s. s = v0t + 1/2at2 = 10m/s × 6s + 1/2 × (-2) m/s2 × (6s) 2 = 24m. Analysis of this method is used blindly Question 6 seconds to apply the formula. This question in the car from brake to stop not 6 seconds, as long as 5 seconds, 1 second is parked in place, the general steps to solve such problems are: (1) Calculate the parking time according to the formula t stop=0-v0/a=-v0/a; (2) compare the size of the time given by the title t with the stop time t stop; (3) according to the size of t and t stop, Determine which of the following methods to use