A particle moves along a straight line in the following way.
starting form the point P0 it moves 6cm to another point P1;then it moves from P1 to the mid-point P2of P0P1;then it moves to the mid-point P3 of P1P2;and the process is repeated indefinitely in a similar way.Find
(a)the total distance that the particle has moves,
(b)the distance between P0and the final position of the particle.
answers:(a)12cm (b)4cm
how can take that??pls
infinity of geometric??
2007-05-30 8:00 pm
回答 (2)
2007-05-30 8:28 pm
✔ 最佳答案
P0P1=6P1P2=P0P1/2=3
P2P3=P1P2/2=1.5
...
It forms a geometric sequence
The first term is 6
The common ratio = 6/3=1/2
a. The total distance =a/(1-r)=6/(1-1/2)=12
b.
The distance between P0 and the final position of the particle
=6-3+1.5-0.75+...
=(6+1.5+...)-(3+0.75+...) [see remark]
=6/(1-1/4)-3/(1-1/4)
=4
Remark:
6, 1.5 , ... form the geometric sequence with first term 6, common ratio (1.5/6=1/4)
3,0.75, ...form the geometric sequence with first term 3, common ratio (0.75/3=1/4)
2007-05-30 12:42:11 補充：
haha ...原來6,-3,1.5,-0.75,...也是一個gp...只要公式一用...6/(1-(-1/2))=6x2/3=4
2007-05-30 8:28 pm
By the use of AP,
a) Distance = 6 + 3 + 1.5 + .....
= 6 X (1 + 0.5 + 0.25 +.....) ( Summation 2^-n , n=0,1,2,...... )
= 6 X 1/(1-0.5)
= 6 X 2
= 12 cm
b) Displacement = 6 - 3 + 1.5 - ......
= 6 X (1 - 0.5 + 0.25 -.....) ( Summation -2^-n , n=0,1,2,...... )
= 6 X 1/(1-(-0.5))
= 6 X 1/1.5
= 4 cm
a) Distance = 6 + 3 + 1.5 + .....
= 6 X (1 + 0.5 + 0.25 +.....) ( Summation 2^-n , n=0,1,2,...... )
= 6 X 1/(1-0.5)
= 6 X 2
= 12 cm
b) Displacement = 6 - 3 + 1.5 - ......
= 6 X (1 - 0.5 + 0.25 -.....) ( Summation -2^-n , n=0,1,2,...... )
= 6 X 1/(1-(-0.5))
= 6 X 1/1.5
= 4 cm
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