When car a and car B start from the same place at the same time, car a decelerates uniformly with the initial speed of 16m / s and the acceleration of 2m / S2; car B accelerates linearly with the initial speed of 4m / s, the acceleration of 1m / S2 and car a in the same direction. The maximum distance between the two cars when they meet again and the movement time of the two cars when they meet again are calculated

When car a and car B start from the same place at the same time, car a decelerates uniformly with the initial speed of 16m / s and the acceleration of 2m / S2; car B accelerates linearly with the initial speed of 4m / s, the acceleration of 1m / S2 and car a in the same direction. The maximum distance between the two cars when they meet again and the movement time of the two cars when they meet again are calculated

When the speed of two cars is equal, the distance between two cars is the largest. According to the velocity formula of uniform linear motion, VT = V A-A a T1, VT = v b + a B T1, the simultaneous solution of the two formulas is t = 4S; at this time, the maximum distance between two cars is △ s = s a − s b = (V a T1 − 12at12) − (v b T1 + 12a B T12) = 24 (m). When B car overtakes a, the motion displacement of two cars is equal, v a t-12a T2 = v b t + 12a B T2, the solution is: T = 2 (V a − v b) a a + a B = 8 (s), (t = 0, rounding off) a: the maximum distance between the two workshops is 24m; when the two cars meet again, the movement time is 8s