Question

物體以一定的初速度從斜面底端A點(diǎn)沖上固定的光滑斜面，斜面總長度為l，到達(dá)斜面最高點(diǎn)C時(shí)速度恰好為零，如圖1.已知物體運(yùn)動(dòng)到距斜面底端l處的B點(diǎn)時(shí)，所用時(shí)間為t，求物體從B滑到C所用的時(shí)間．

圖1

Answer 1

解析　方法一：逆向思維法

物體向上勻減速?zèng)_上斜面，相當(dāng)于向下勻加速滑下斜面．

故x_BC＝，x_AC＝，又x_BC＝，

由以上三式解得t_BC＝t.

方法二：基本公式法

因?yàn)槲矬w沿斜面向上做勻減速運(yùn)動(dòng)，設(shè)物體從B滑到C所用的時(shí)間為t_BC，由勻變速直線運(yùn)動(dòng)的規(guī)律可得

v＝2ax_AC                                                                                                                                                                                                                       ①

v＝v－2ax_AB                                                                                                                                                                                                               ②

x_AB＝x_AC                                                                                                                                                                                                                       ③

由①②③解得v_B＝                                                                                            ④

又v_B＝v₀－at                                                                                                          ⑤

v_B＝at_BC                                                                                                                                                                                                                                   ⑥

由④⑤⑥解得t_BC＝t.

方法三：比例法

對于初速度為零的勻加速直線運(yùn)動(dòng)，在連續(xù)相等的時(shí)間里通過的位移之比為x₁∶x₂∶x₃∶…∶x_n＝1∶3∶5∶…∶(2n－1)．

因?yàn)?i>x_CB∶x_BA＝∶＝1∶3，而通過x_BA的時(shí)間為t，所以通過x_BC的時(shí)間t_BC＝t.

方法四：中間時(shí)刻速度法

利用推論：勻變速直線運(yùn)動(dòng)中中間時(shí)刻的瞬時(shí)速度等于這段位移的平均速度，_AC＝＝.又v＝2ax_AC，v＝2ax_BC，x_BC＝.由以上三式解得v_B＝.可以看成v_B正好等于AC段的平均速度，因此B點(diǎn)是這段位移的中間時(shí)刻，因此有t_BC＝t.

方法五：圖象法

根據(jù)勻變速直線運(yùn)動(dòng)的規(guī)律，畫出v－t圖象．如圖所示．利用相似三角形的規(guī)律，面積之比等于對應(yīng)邊的平方比，得＝，且＝，OD＝t，OC＝t＋t_BC.所以＝，解得t_BC＝t.

答案　t