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问题 问答题

一火车以2m/s的初速度，0.6m/s2的加速度做匀加速直线运动，求：

（1）火车在第4s末的速度是多少？

（2）在第6s内的位移是多少？

答案

（1）由题意可知：v0=2m/s，a=0.6m/s2，由v=v0+at 

可得4s末的速度：v=2+0.6×4=4.4m/s

（2）火车在前6s内的位移为：x6=v0t6+

1
2
a
t26

  代入数据得：x6=2×6+

1
2
×0.6×62m=22.8m

火车在前5s内的位移为：x5=v0t5+

1
2
a
t25

代入数据得：x5=17.5m

所以，火车在第6s内的位移为△x=x6-x5=5.3m

答：

（1）火车在第4s末的速度是4.4m/s．

（2）在第6s内的位移是17.5m．

填空题

客专隧道模板台车应保证台车走行轨道中线和隧道中线重合，模板台车走行轨道的中线和标高误差应不大于（）mm。
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单项选择题

A transport layer protocol usually has several responsibilities．One is to create a process-to-process communication；UDP uses (1) numbers to accomplish this．Another responsibility is to provide control mechanisms at the transport level．UDP does this task at a very minimal level．There is no flow control mechanism and there is no(2) for received packet．UDP, however, does provide error control to some extent．If UDP detects an error in the received packet，it will silently drop it． The transport layer also provides a connection mechanism for the processes．The (3) must be able to send streams of data to the transport layer．It is the responsibility of the transport layer at (4) station to make the connection with the receiver, chop the stream into transportable units，number them，and send them one by one．It is the responsibility of the transport layer at the receiving end to wait until all the different units belonging to the same process have arrived，check and pass those that are (5) free, and deliver them to the receiving process as a stream．

（2）处应填（）。

A．connection

B．window

C．acknowledgement

D．destination
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