For example, car A and car B are 4 meters apart, the initial speed of car B is 0, the acceleration is 1 m/s 2, and car A travels at a constant speed of 3 meters/s,

Then according to the calculation 1/2 * 1 * t 2+4 = 3t,

Simplify it to t 2-6t+8 = 0, and the solution is t=2 or t=4.

That is to say, two seconds later, vehicles A and B meet. At this time, vehicle B is driving at a distance of 2m+4m = 6m from the starting point of vehicle A. ..

Car A travels 3*2=6 meters to verify the encounter.

At this time, the speed of car B is 2m/s, which is slower than that of car A, and car A overtook it.

After 4 seconds, the driving distance between car B and the starting point of car A is1/2 *1* 4 2+4 = 8m+4m =12m.

Car A travels 3*4= 12 meters, and car A meets car B..

At this time, the speed of B car is 4m/s, which is faster than A car, and B car has surpassed it.

So, it's not that car A slowed down after overtaking, but that when it caught up for the first time, car A was fast and car B hadn't come up yet.

The second time, the second car that had fallen behind came, and the second car caught up again.