选D;
x1=(1/2)a[t^2]
x2=(1/2)a[(t+1)^2]
t=1s
联立方程组,解得
x=x2—x1
=(1/2)a[(t+1)^2]—(1/2)a[t^2]
=(1/2)×a×2×2—(1/2)×a×1×1
=3a/2
故a=2x/3
选D;
x1=(1/2)a[t^2]
x2=(1/2)a[(t+1)^2]
t=1s
联立方程组,解得
x=x2—x1
=(1/2)a[(t+1)^2]—(1/2)a[t^2]
=(1/2)×a×2×2—(1/2)×a×1×1
=3a/2
故a=2x/3