| v甲 |
| t1 |
| 30 |
| 12 |
a乙m=
| v乙 |
| t2 |
| 30 |
| 6 |
加速到最大速度需要的时间:t甲m=
| 40 |
| 2.5 |
t乙m=
| 50 |
| 5 |
到相遇时乙行驶的距离为x乙=
| 1 |
| 2 |
| 1 |
| 2 |
甲加速阶段行驶的最大距离为:x甲m=
| 1 |
| 2 |
| 1 |
| 2 |
因为x乙m+85<x甲m,所以,在甲乙加速阶段相遇.
由
| 1 |
| 2 |
| 1 |
| 2 |
即
| 1 |
| 2 |
| 1 |
| 2 |
解得:t0=6s
(2)两车相遇时甲车行驶的路程为:x甲=
| 1 |
| 2 |
| 1 |
| 2 |
答:(1)t0应该满足的条件是6s.
(2)两车相遇时甲车行驶的路程是245m.
| 起动的快慢/s (0~30m/s的加速时间) | 最大速度/m•s-1 | |
| 甲车 | 12 | 40 |
| 乙车 | 6 | 50 |
| v甲 |
| t1 |
| 30 |
| 12 |
| v乙 |
| t2 |
| 30 |
| 6 |
| 40 |
| 2.5 |
| 50 |
| 5 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |