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? <<<

- anonymous

- chestercat

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- anonymous

- hartnn

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

use the same formula as in part a)

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## More answers

- anonymous

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

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 \(\large a_1 \dfrac{1-r^n}{1-r}\)

- anonymous

we need to find n.
r=2/3. how about a1?

- anonymous

- anonymous

a1=2m

- anonymous

ohh yes

- anonymous

but..

- anonymous

whats the problem?

- anonymous

(2/3)^n = 3/25 ?

- anonymous

then use log

- anonymous

...wait

- anonymous

5.23??

- anonymous

5.8 = {2(1-(2/3)^n)}/(1-(2/3))

- anonymous

did u solve for n?

- anonymous

n=5.23??

- anonymous

\[2(\frac{ 1-(\frac{ 2 }{ 3 } )^{n}}{ 1-(\frac{ 2 }{ 3 }) })=5.28\]

- anonymous

no...... it is 5.8

- anonymous

yaya, typing mistake

- anonymous

u used 5.28

- anonymous

I got 8.39

- anonymous

8.388

- anonymous

got it now, thanks

- anonymous

welcome

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