marilyn cycles at an average speed of 10.5ms-1 around a circular track in anticlockwise direction from A(i.e.the north of the circular track). She stops after 3 minutes. The radius of the circular track is 400m.
Q:what is her average velocity?
f.4physics questioin(velocity)
2012-09-21 2:47 am
回答 (3)
2012-09-25 12:16 am
✔ 最佳答案
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參考: miraco
2012-09-24 6:51 am
The total diplacement of marilyn should be √(400^2+400^2) =565m.
velocity=total displacement / total time taken
velocity=565 / 180
Therefore, her average velocity = 3.14 ms-1.
The angle should be 90 / 2 =45度(isoceles triangle)，which is south 45度east
from the starting point.
Therefore, her total average velocity = 3.14ms-1 (S 45度E)
velocity=total displacement / total time taken
velocity=565 / 180
Therefore, her average velocity = 3.14 ms-1.
The angle should be 90 / 2 =45度(isoceles triangle)，which is south 45度east
from the starting point.
Therefore, her total average velocity = 3.14ms-1 (S 45度E)
參考: me
2012-09-21 5:26 am
Total distance
=10.5*3*60
=1890m
Let a rad be the central angle corresponding with the arc(the 1890m that marilyn rode.)
1890 / 2pi(400) = a / 2pi <-------係in radian
a=1890 / 400
=4.725
Total displacement
=√[400^2 + 400^2 -2(400)(400)cos(2pi rad -4.725 rad)]
=√(3200-3200cos 4.725 rad) m
=√[3200(1-cos 4.725 rad)] m
=40√2[1-cos 4.725 rad)] m
Average velocity
=40√2[1-cos 4.725 rad)] / 180
=0.312 m s^-1
2012-09-20 22:31:00 補充：
1rad / 2pi rad = 1度 / 360度
咁Let a degree(度 5識打)
1890 / 2pi(400) = a / 360
a= 189*9 / 2pi
之後total displacement
cos(2pi rad -4.725 rad)] 轉做
cos(360 - 189*9 / 2pi )
=cos (189*9 / 2pi)
咁樣轉晒佢就得了
p.s.如果5識cosine theorem既話 我就5識做啦.... 應該俾用掛
=10.5*3*60
=1890m
Let a rad be the central angle corresponding with the arc(the 1890m that marilyn rode.)
1890 / 2pi(400) = a / 2pi <-------係in radian
a=1890 / 400
=4.725
Total displacement
=√[400^2 + 400^2 -2(400)(400)cos(2pi rad -4.725 rad)]
=√(3200-3200cos 4.725 rad) m
=√[3200(1-cos 4.725 rad)] m
=40√2[1-cos 4.725 rad)] m
Average velocity
=40√2[1-cos 4.725 rad)] / 180
=0.312 m s^-1
2012-09-20 22:31:00 補充：
1rad / 2pi rad = 1度 / 360度
咁Let a degree(度 5識打)
1890 / 2pi(400) = a / 360
a= 189*9 / 2pi
之後total displacement
cos(2pi rad -4.725 rad)] 轉做
cos(360 - 189*9 / 2pi )
=cos (189*9 / 2pi)
咁樣轉晒佢就得了
p.s.如果5識cosine theorem既話 我就5識做啦.... 應該俾用掛
收錄日期: 2021-04-13 18:59:18
原文連結 [永久失效]:
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