## 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 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? <<<

1. anonymous

@hartnn

2. 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

3. hartnn

use the same formula as in part a)

4. 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

5. 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 $$\large a_1 \dfrac{1-r^n}{1-r}$$

6. anonymous

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

7. anonymous

@sauravshakya

8. anonymous

a1=2m

9. anonymous

ohh yes

10. anonymous

but..

11. anonymous

whats the problem?

12. anonymous

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

13. anonymous

then use log

14. anonymous

...wait

15. anonymous

5.23??

16. anonymous

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

17. anonymous

did u solve for n?

18. anonymous

n=5.23??

19. anonymous

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

20. anonymous

no...... it is 5.8

21. anonymous

yaya, typing mistake

22. anonymous

u used 5.28

23. anonymous

I got 8.39

24. anonymous

8.388

25. anonymous

got it now, thanks

26. anonymous

welcome