题目内容

In the following graph,the velocity of an object as it moves along a horizontal straight line(水平直线)is plotted against time.(一个物体在水平直线上运动,其速度随时间的变化如图所示)
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(1)The distance traveled by the object during the frist 6seconds in
 
meters.(物体在前6s内的位移是多少)
(2)The average speed of the object during the first 8seconds is in meters per second
 
.(物体在前8s内的平均速率是多少)
分析:(1)速度时间图象与时间轴所围的“面积”表示位移,由几何知识求解位移;
(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)速度时间图象与时间轴所围的“面积”表示位移,则物体在前6s内的位移是
  x=
1
2
×(2+6)×4m=16m
(2)物体在前8s内的位移为:
x′=
1
2
×
(2+8)×4=20m
由于前8s内物体做单向直线运动,路程等于位移的大小,则前8s内路程为S=x′=20m
则平均速率
.
v
=
S
t
=
20
8
=2.5m/s
故答案为:16,2.5m/s.
点评:解答本题关键掌握速度图象的物理意义:面积表示位移,以及平均速率公式.
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