Question

integral 1/-32-.015v dv= integral dt

Answer 1

1/(-32-.015v) dv

Answer 2

\[\int\frac1{-32-0.015v}dv\]this?

Answer 3

\[u=-32-0.015v\implies du=?\]

Answer 4

-.015 dv

Answer 5

good, so\[du=-0.015dv\implies dv=?\]

Answer 6

du/-.015

Answer 7

great now plug that into the original intagral

Answer 8

\[\int\limits_{?}^{?} 1/u du=\]

Answer 9

-.015ln(-32-.015v)

Answer 10

you are missing a constant....

Answer 11

not quite, but close

Answer 12

help please

Answer 13

what was du again? you said it earlier...

Answer 14

you are very close, but I want you to see your mistake

Answer 15

-.015dv

Answer 16

sorry I mean what was dv ?
you said it earlier...

Answer 17

yes, so dv=....?

Answer 18

-1/.015 du

Answer 19

good, so where did that -1/0.015 go when you made your substitution?

Answer 20

you dropped it ;) gotta be a bit more careful there

Answer 21

brb, try again now with that tip

Answer 22

\[\int\limits_{?}^{?} 1/u (-1/.015)du=-1/.015 \ln u\]

Answer 23

..+C
yes, now just sub back in for u :)

Answer 24

heres the problem:
A woman bails out of an airplane at altitude of 10,000 ft falls freely for 20s, then opens parachute. How long will it take her to reach the ground? ssume linear air resistance pv ft/s^2, taking p=.15 without the parachute and p=1.5 with the parachute. (first determine her height aabove ground and velocity when the parachute opens)

accelercation due to gravity would be -32 ft/s^2
so I
dv/dt--32-.15v v<0
\[\int\limits_{?}^{?}1/(-32-.15v) dv=\int\limits_{?}^{?} dt\]
give me
-1/.15 ln(-32-.15t)= t+C
v(0)= -1/.15 ln(-32-.15(0))=0+C
c=??

since it says linear can I
v=e^integral p = e^-.15t(integral -32e^-.15t)= e^-.15t(-32/.15 e^.15t+C)

look at a solution and I do not follow so can you walk me through this?

Answer 25

well now this is a word problem, so let me have lunch and I will get back to you on this, okay?

Answer 26

thank you