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
設(shè)加速度為 a ,總時(shí)間 為 t
則：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
帶入 ① at - 3a = 0.4 得：
5a - 3a = 0.4
2a = 0.4
a = 0.2 m/s²滑到斜面地端時(shí)的速度就是V = at = 0.2x5 = 1m/s