kryton1212
A snake starts moving at a speed of 2 m per second towards a target which is 5.8m away. In each second, the snake only moves two-thirds of the distance it traveled in the previous second.
(a) Find the total distance that the snake has moves in the first five seconds. [Done; 5.21m]
(b) Find the total distance that the snake can travel. [Done; 6m]
>>>(c) When will the snake reach the target? <<<
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kryton1212
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@hartnn
hartnn
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the sum of distance, total distance = 5.8
so, plug in S= 5.8 in the equations you used, and keep time 't' as t only
hartnn
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use the same formula as in part a)
kryton1212
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in part a, i work like this:
2+2*[(2/3)+(2/3)^2+...+(2/3)^4]
=2+2*{[(2/3)(1-(2/3)^4]/[1-(2/3)]}
=5.21
hartnn
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so, for part c, you need to find number of terms, right ? let that be n
2+2* [(2/3)+(2/3)^2+....(2/3)^n]
(2/3)+(2/3)^2+....(2/3)^n is the geometric series
a1 = 2/3, r=2/3
so, instead of (2/3)+(2/3)^2+....(2/3)^n
write \(\large a_1 \dfrac{1-r^n}{1-r}\)
kryton1212
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we need to find n.
r=2/3. how about a1?
kryton1212
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@sauravshakya
sauravshakya
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a1=2m
kryton1212
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ohh yes
kryton1212
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but..
sauravshakya
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whats the problem?
kryton1212
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(2/3)^n = 3/25 ?
sauravshakya
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then use log
kryton1212
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...wait
kryton1212
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5.23??
sauravshakya
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5.8 = {2(1-(2/3)^n)}/(1-(2/3))
sauravshakya
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did u solve for n?
kryton1212
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n=5.23??
kryton1212
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\[2(\frac{ 1-(\frac{ 2 }{ 3 } )^{n}}{ 1-(\frac{ 2 }{ 3 }) })=5.28\]
sauravshakya
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no...... it is 5.8
kryton1212
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yaya, typing mistake
sauravshakya
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u used 5.28
sauravshakya
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I got 8.39
kryton1212
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8.388
kryton1212
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got it now, thanks
sauravshakya
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welcome