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anonymous
 3 years ago
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 twothirds 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? <<<
anonymous
 3 years ago
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 twothirds 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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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1the 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
 3 years ago
Best ResponseYou've already chosen the best response.1use the same formula as in part a)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in 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
 3 years ago
Best ResponseYou've already chosen the best response.1so, 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 \(\large a_1 \dfrac{1r^n}{1r}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we need to find n. r=2/3. how about a1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05.8 = {2(1(2/3)^n)}/(1(2/3))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[2(\frac{ 1(\frac{ 2 }{ 3 } )^{n}}{ 1(\frac{ 2 }{ 3 }) })=5.28\]
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