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# 江苏专用2019版高考物理大一轮复习第一章运动的描述第3讲运动的图象追及和相遇问题

2019 3

1 xt 1 x t( )
1 A B C D xt xt xt xt B ACD B 2 v d a1 a2 ( ) A B C v> a1a2 d D v< a1a2 d d v0 v202(a1a2)d v>v0 v<v0 D 3 vt 2 ( )

2 A t1 B t1 C t1 D t1 t1 AB D C AB 4 3 1234 ( )
3 A 1 Bxt t1 v1>v2 Cvt 0 t3 3 4 D 2 4 t2t4 xt vt A xt t1 v1>v2B vt t 0t3 3 4 3 4 C 2 t2 4 t4 D B 5 4 4 vt ( )
4

at 01 s 1 m/s2 0 1 s 1 m/s,12 s 1 m/s2 1 m/s 2 s 0 C. C 6 t1 v1 t2 v2 5

0t1 t1t2 v( )

5

A0t1vv21

Bt1t2vv12 v2

Ct1t2v>v12 v2

Dt1t2v<v12 v2

0t1 v02 v1v21 A t1t2

( vt t

)v<v12 v2 BC D

7 6

( )

6 A 80 s B70 s C60 s D80 s 80 s A 70 s B 80 s C D D 8 7 vt xt ( x )( )
7
0t1 A t1t2 BD t2t3 xt 0t1 t2t3 C C 9 vt 8 (F x )( )

8
02 s 2 s 24 s 46 s 68 s A B 4 s CD B 10 4 m v0 x v20( xv20) 9
9 (1) (2) 5.0 m/s (1) v202ax ""21a a2 m/s2. (2) 4 m/s t xv0t12at245t122t2 t1 st4 s( ) (1)2 m/s2 (2)1 s

11 10 m/s 2 s 2 m/s2 (1) (2) vm12 m/s (1) t1 t1120 s5 s x (25)10 m70 m x 12at2112252 m25 m xx x 45 m. (2) t2122 s6 sx (26)10 m80 m x 12at2212262 m36 m x >x xx x 44 m t12t4410t t22 st 28 s. (1)45 m (2)28 s
12 AB A vA10 m/sB vB30 m/s.B A 600 m A B B 1 800 m (1)B (2) B 8 s A a10.5 m/s2 (1) B a 0v2B2ax1 a0.25 m/s2. (2) B t vBatvAa1(t t) t32 s at2 B xBvBt 2 832 m A xAvA tvA(t t)12a1(t t)2464 m

xAx>xB xxAxxB232 m. (1)0.25 m/s2 (2) 232 m