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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? <<<

Mathematics
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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
use the same formula as in part a)

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Other answers:

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
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 \(\large a_1 \dfrac{1-r^n}{1-r}\)
we need to find n. r=2/3. how about a1?
a1=2m
ohh yes
but..
whats the problem?
(2/3)^n = 3/25 ?
then use log
...wait
5.23??
5.8 = {2(1-(2/3)^n)}/(1-(2/3))
did u solve for n?
n=5.23??
\[2(\frac{ 1-(\frac{ 2 }{ 3 } )^{n}}{ 1-(\frac{ 2 }{ 3 }) })=5.28\]
no...... it is 5.8
yaya, typing mistake
u used 5.28
I got 8.39
8.388
got it now, thanks
welcome

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