The object starts from the rest and moves in a straight line with uniform acceleration of A1. After a period of time, it moves in a straight line with uniform deceleration of A2 to the rest. If the total displacement is x, the maximum velocity in the motion is () A. 2a1xa2B. (a1+a2)xC. （a1+a2）xD. 2a1a2xa1+a2

Let the first acceleration time be T1, and the second acceleration time be T2. The maximum velocity of the acceleration process is v = a1t1 & nbsp; & nbsp; & nbsp; deceleration to zero. We can see its inverse process, which is equivalent to the uniformly accelerated motion with zero initial velocity, and get v = a2t2 & nbsp; from the above two formulas: T2 = a1t1a2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① so the whole displacement: x = V2 (T1 + T2) & nbsp; & nbsp; ② From the solution of (1) and (2), we find that: T1 = 2a2xa1a2 + A21, so: v = a1t1 = 2a1a2x (a1 + A2), so: D

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