解:(1)設(shè)運(yùn)動(dòng)t秒時(shí),BC=8單位長度,
①當(dāng)點(diǎn)B在點(diǎn)C的左邊時(shí),
由題意得:6t+8+2t=24,
解得:t=2(秒);
②當(dāng)點(diǎn)B在點(diǎn)C的右邊時(shí),
由題意得:6t-8+2t=24,
解得:t=4(秒).
(2)當(dāng)運(yùn)動(dòng)2秒時(shí),點(diǎn)B在數(shù)軸上表示的數(shù)是4;
當(dāng)運(yùn)動(dòng)4秒時(shí),點(diǎn)B在數(shù)軸上表示的數(shù)是16.
(3)存在關(guān)系式
=3.
設(shè)運(yùn)動(dòng)時(shí)間為t秒,
1)當(dāng)t=3時(shí),點(diǎn)B和點(diǎn)C重合,點(diǎn)P在線段AB上,0<PC≤2,且BD=CD-4,AP+3PC=AB+2PC=2+2PC,
當(dāng)PC=1時(shí),BD=AP+3PC,即
=3;
2)當(dāng)3<t<
時(shí),點(diǎn)C在點(diǎn)A和點(diǎn)B之間,0<PC<2,
①點(diǎn)P在線段AC上時(shí),BD=CD-BC=4-BC,AP+3PC=AC+2PC=AB-BC+2PC=2-BC+2PC,
當(dāng)PC=1時(shí),有BD=AP+3PC,即
=3;
點(diǎn)P在線段AC上時(shí),BD=CD-BC=4-BC,AP+3PC=AC+4PC=AB-BC+4PC=2-BC+4PC,
當(dāng)PC=
時(shí),有BD=AP+3PC,即
=3;
3°當(dāng)t=
時(shí),點(diǎn)A與點(diǎn)C重合,0<PC≤2,BD=CD-AB=2,AP+3PC=4PC,
當(dāng)PC=
時(shí),有BD=AP+3PC,即
=3;
4°當(dāng)
<t
時(shí),0<PC<4,BD=CD-BC=4-BC,AP+3PC=AB-BC+4PC=2-BC+4PC,
PC=
時(shí),有BD=AP+3PC,即
=3.
分析:(1)設(shè)運(yùn)動(dòng)t秒時(shí),BC=8(單位長度),然后分點(diǎn)B在點(diǎn)C的左邊和右邊兩種情況,根據(jù)題意列出方程求解即可;
(2)由(1)中求出的運(yùn)動(dòng)時(shí)間即可求出點(diǎn)B在數(shù)軸上表示的數(shù);
(3)隨著點(diǎn)B的運(yùn)動(dòng),分別討論當(dāng)點(diǎn)B和點(diǎn)C重合、點(diǎn)C在點(diǎn)A和B之間及點(diǎn)A與點(diǎn)C重合時(shí)的情況.
點(diǎn)評(píng):本題考查兩點(diǎn)間的距離,并綜合了數(shù)軸、一元一次方程和線段長短的比較,難度較大,注意對(duì)第三問進(jìn)行分情況討論,不要漏解.