Car a starts to move in a straight line with a constant acceleration of 2 meters every second. Car B lags behind by 4 seconds at the same place The two cars are moving in the same direction. Ask 1: what is the maximum distance between the two cars before car B overtakes car a? 2: how long does it take for car B to catch up with car a after starting? How far are they from the starting point at this time?

Car a starts to move in a straight line with a constant acceleration of 2 meters every second. Car B lags behind by 4 seconds at the same place The two cars are moving in the same direction. Ask 1: what is the maximum distance between the two cars before car B overtakes car a? 2: how long does it take for car B to catch up with car a after starting? How far are they from the starting point at this time?

1. Because only when the speed of two cars is equal can there be the maximum distance,
At this time, the time of vehicle B is t,
2*（t+4）=8*t
T = 4 / 3A parking space shift s = 1 / 2 * 2 * 256 / 9 = 256 / 9 B parking space shift s = 1 / 2 * 8 * 16 / 9 = 32 / 9
The maximum distance is 224 / 9
2. Let t be the time to catch up with car a‘
At this time, the displacement of car a S1 = 2 * 1 / 2 * (t '+ 4) * (t' + 4)
Vehicle B displacement S2 = 1 / 2 * 8 * t '* t'
The displacement of car a in 4 seconds is s = 2 * 1 / 2 * 4 * 4 = 16
s2=s1+6
Find out t, which can be solved later