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?

- anonymous

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- anonymous

I have t=6

- anonymous

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

- anonymous

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

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## More answers

- anonymous

Grrrrrrr Idk HHHHEEEEELLLLLPPPPPP

- anonymous

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

- anonymous

yes i get 6 also

- anonymous

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

- anonymous

he is correct

- anonymous

sorry, u have made displacement 12 :)

- anonymous

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?

- anonymous

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

- anonymous

total distance covered, i guess

- anonymous

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).

- anonymous

Ktklown the total distance

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

yeah,@ ktklown, u r right :)

- anonymous

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

- anonymous

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)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

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