In the following graph，the velocity of an object as it moves along a horizontal straight line（水平直線）is plotted against time．（一個物體在水平直線上運動，其速度隨時間的變化如圖所示）
精英家教網(wǎng)
（1）The distance traveled by the object during the frist 6seconds in
 
meters．（物體在前6s內(nèi)的位移是多少）
（2）The average speed of the object during the first 8seconds is in meters per second
 
．（物體在前8s內(nèi)的平均速率是多少）
分析：（1）速度時間圖象與時間軸所圍的“面積”表示位移，由幾何知識求解位移；
（2）先求出位移，再由平均速率公式
.
v
=
S
t
求平均速率．
解答：解：（1）速度時間圖象與時間軸所圍的“面積”表示位移，則物體在前6s內(nèi)的位移是
  x=
1
2
×（2+6）×4m=16m
（2）物體在前8s內(nèi)的位移為：
x′=
1
2
×（2+8）×4=20m
由于前8s內(nèi)物體做單向直線運動，路程等于位移的大小，則前8s內(nèi)路程為S=x′=20m
則平均速率
.
v
=
S
t
=
20
8
=2.5m/s
故答案為：16，2.5m/s．
點評：解答本題關(guān)鍵掌握速度圖象的物理意義：面積表示位移，以及平均速率公式．
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