## patbatE21 3 years ago Calculus II: Starting at s = 0 when t = 0, an object moves along a line so that its velocity at time t is v(t) = 2t - 4 centimeters per second. How long will it take to get to s = 12? To travel a total distance of 12 centimeters?

1. patbatE21

I have t=6

2. ktklown

hold on, i'll compute and see if it matches

3. patbatE21

v(t) = 2t - 4 s(t) = integral 2t - 4 = 2t^2 - 4t + C 0 = 0^2 -4(0) + C t^2 - 4t = 12 t^2 -4t -12=0 (t-6)(t-2) t= 6 considering time is never negative

4. patbatE21

Grrrrrrr Idk HHHHEEEEELLLLLPPPPPP

5. Arnab09

that will be (t-6)(t+2)

6. ktklown

yes i get 6 also

7. Arnab09

second thing, u have made the displacement 0, but actually distance is needed

8. patbatE21

he is correct

9. Arnab09

sorry, u have made displacement 12 :)

10. ktklown

yes since the velocity starts out negative, the object is going backwards at first, then goes forwards. do you want the total displacement to be 12 or the position from the starting point to be 12?

11. patbatE21

Arnab09 would that mean I would have to integrate as such|dw:1321863723148:dw|

12. Arnab09

total distance covered, i guess

13. ktklown

if you want the total displacement to be 12, then you first solve to find out where the object stopped (0 velocity) by setting v(t)=0. then plug in that t value to s(t).

14. patbatE21

Ktklown the total distance

15. ktklown

OK, so first solve to see where the object changed direction. that happened when v(t)==0, i.e. at t=2

16. patbatE21

ok

17. ktklown

at t=2 the object was at -4. then it starts going forward, so its total displacement will equal 12 when it reaches +4

18. ktklown

i.e. from start (0) to -4, then from -4 back to 0 for a total of 8, then from 0 to +4 for a total of 12

19. patbatE21

ok

20. ktklown

it looks like that happens at 2+2sqrt(3) =t

21. Arnab09

yeah,@ ktklown, u r right :)

22. Arnab09

put the position 8 inspite of 12, u will get the answer

23. ktklown

It looks like, from reading the original question, they want to know both answers. First when does it reach s=12, and second when does it travel a total of 12 (in which case we do the other computation we just finished)

24. Arnab09

in the question, the term is 'distance', not 'displacement' so the second method is correct, definitely..

25. patbatE21

I've worked it out 4 times today, I'm probably just bad at this

26. ktklown

well it specifically says "s=12", so it seems to me like they are asking both versions. i don't know, it's not very clearly worded

27. Arnab09

maybe, if s (displacement) =12, then t^2-4t=12 and t=6

28. ktklown

yeah, if i were doing this as homework i'd probably write the answer for both interpretations