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)

(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)
(1)The distance traveled by the object during the frist 6seconds in
(2)The average speed of the object during the first 8seconds is in meters per second
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
=
求平均速率.
(2)先求出位移,再由平均速率公式
. |
| v |
| S |
| t |
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=
×(2+6)×4m=16m
(2)物体在前8s内的位移为:
x′=
×(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
=
=
=2.5m/s
故答案为:16,2.5m/s.
x=
| 1 |
| 2 |
(2)物体在前8s内的位移为:
x′=
| 1 |
| 2 |
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 |
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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