(1)第二秒的位移,[f(2)-f(1) ]/ 1 = -0.5
(2)第一秒末速度 f'(t) = 9t-6t^2
, f'(1) = 3, f'(2) = -6
及速度 f''(t) = 9-12t
f''(1) = -3, f''(2) = -15
(3)第二秒内的路程,求x=f(t)最值,f'(t) = 9t-6t^2 = 0, t = 1.5
因此物体从f(1) 运动到f(1.5)再回到f(2)
f(1) = 2.5, f(1.5) = 27/8, f(2) = 2
路程 = (27/8 - 2.5) +(27/8 - 2)=9/4
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