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jennychan12
Group Title
AP Calculus AB 2003 #2
http://www.collegeboard.com/prod_downloads/ap/students/calculus/calculus_ab_frq_03.pdf
I don't understand how to do part c).
 one year ago
 one year ago
jennychan12 Group Title
AP Calculus AB 2003 #2 http://www.collegeboard.com/prod_downloads/ap/students/calculus/calculus_ab_frq_03.pdf I don't understand how to do part c).
 one year ago
 one year ago

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bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
total distance is integral of velocity but u gotta take into consideration which way it's going
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
I looked at the sample respoinses and they did like absolute value of v(t). But I don't understand why. http://apcentral.collegeboard.com/apc/public/repository/ap03_calculusab_q2_27980.pdf
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
so you found where it changes direction right?
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
the thing is...dw:1357322365854:dw
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
if u take just the integral, the area on the bottom gets subtracted
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
but that's for displacement. So if u want to get the distance you need to add that area instead of subtracting it.. So you have to integrate from start till when it changed direction, then from when it changed direction till when it changed direction again, etc.. and finally take the absolute value of all integrals and add them up
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
dw:1357322510296:dw see if u start at pt A, displacement, which is the regular integral, is 0, but distance traveled, is 5
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
i thought to find the distance u have to take integral of the velocity.
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
yes you do.. integral of velocity is indeed the area under the velocity curve, which is the displacement
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
so in order to get the DISTANCE.. you want to add all the areas that end up below y axis, instead of subtracting them
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
if you were looking for DISPLACEMENT, you would just take the integral of that function from 0 to 3. But, since you're looking for DISTANCE, take the integral of that function from 0 to a, where a is the time at which particle changed direction.
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
okay listen, where did the particle change direction?
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
rad(2pi) and 0
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
yeah those were the solutions but only rad(2pi) works
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
no it was just t = 2.507
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
yeah that's rad(2pi)
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
so integrate that function from 0 to 2.507, then from 2.507 to 3
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
once you get the results, say Int from 0 to 2.507 = A, and Int from 2.507 to 3 = B; just add up the absolute values: A + B
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
that's all he's doing by integrating the absolute value of V over 0 < t < 3
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
how'd u integrate the function tho? like algebraically?
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
isn't that the calculator part?
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
i dont think u can integrate exponents inside trig functions yet.. lol
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
or do u just plug it into the graphing calculator? ohhh _ ok.
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
i mean i dont think there is a way that anyone has found how to do that
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
lol yea
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
ohhh yeah i got the answer when i plugged it into the calculator
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
You mean to tell me.... I WASTED MY TIME???!!!!!!
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
nooooooo...... like after you told me all that...
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.2
k good lol
 one year ago

jennychan12 Group TitleBest ResponseYou've already chosen the best response.1
sorry..if anyone's still viewing this for part d) why did they add 1 to rad(2pi) ?
 one year ago
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