Question

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? <<<

Answer 1

@hartnn

Answer 2

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

Answer 3

use the same formula as in part a)

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

@sauravshakya

Answer 8

a1=2m

Answer 9

ohh yes

Answer 10

but..

Answer 11

whats the problem?

Answer 12

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

Answer 13

then use log

Answer 14

...wait

Answer 15

5.23??

Answer 16

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

Answer 17

did u solve for n?

Answer 18

n=5.23??

Answer 19

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

Answer 20

no...... it is 5.8

Answer 21

yaya, typing mistake

Answer 22

u used 5.28

Answer 23

I got 8.39

Answer 24

8.388

Answer 25

got it now, thanks

Answer 26

welcome