一物从斜面顶端静开始匀加速下滑到斜面底端,初3s经路程S1,末3s经路程,S2,S2-s1=1.2,s1:s2=3:7,
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初速度 Vo = 0

初三秒位移S1

末三秒位移S2

S2 - S1 = 1.2m

S1 :S2 =3 :7

设加速度为 a ,总时间 为 t

则:S1 = 0.5xax3² = 4.5a

最后末速度即:

末三秒末速度 V = at

末三秒初速度 V1 = V - 3a = at -3a

末三秒位移 S2 = 平均速度x3s = 0.5(V1 + V) x 3 = 0.5x(at -3a +at)x3 = 3at - 4.5a

S1 = 4.5a

S2 = 3at - 4.5a

S2 - S1 = 1.2m

即:3at - 4.5a - 4.5a = 1.2

3at - 9a = 1.2

at - 3a = 0.4 ---------①

S1 :S2 = 3 :7

即:4.5a / (3at - 4.5a) = 3 / 7

4.5/(3t -4.5) = 3 / 7

4.5x7 = (3t -4.5)x3

31.5 = 9t - 13.5

9t = 45

t = 5 秒

带入 ① at - 3a = 0.4 得:

5a - 3a = 0.4

2a = 0.4

a = 0.2 m/s²