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?

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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?

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I have t=6
hold on, i'll compute and see if it matches
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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Grrrrrrr Idk HHHHEEEEELLLLLPPPPPP
that will be (t-6)(t+2)
yes i get 6 also
second thing, u have made the displacement 0, but actually distance is needed
he is correct
sorry, u have made displacement 12 :)
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?
Arnab09 would that mean I would have to integrate as such|dw:1321863723148:dw|
total distance covered, i guess
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).
Ktklown the total distance
OK, so first solve to see where the object changed direction. that happened when v(t)==0, i.e. at t=2
ok
at t=2 the object was at -4. then it starts going forward, so its total displacement will equal 12 when it reaches +4
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
ok
it looks like that happens at 2+2sqrt(3) =t
yeah,@ ktklown, u r right :)
put the position 8 inspite of 12, u will get the answer
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)
in the question, the term is 'distance', not 'displacement' so the second method is correct, definitely..
I've worked it out 4 times today, I'm probably just bad at this
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
maybe, if s (displacement) =12, then t^2-4t=12 and t=6
yeah, if i were doing this as homework i'd probably write the answer for both interpretations

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