Question

Anti derivatives. Hewp.

Answer 1

@amistre64 you free? :)

Answer 2

what is a linear expression?

Answer 3

hm? o.o

Answer 4

[oh for the second one under position I got: -16(t-3)^2+297(t-3)+297 which is correct] I just need the last on on the velocity and the last two on position D:

Answer 5

how did you figure out the others on velocity?

Answer 6

did current velocity and displacement haha probably a lot more complicated than I had to make it.

Answer 7

I pretty much guessed at how to do it. I don't really understand it that well.

Answer 8

given an acceleration, we can find an antiderivative to define velocity

knowing velocity, we can find an antiderivative to define position

Answer 9

if a = 66t, then v = 66t^2/2 + c

Answer 10

and c was 0 in that case.

Answer 11

that does appear to be the case

Answer 12

after the fuel runs out, gravity is accelerating it and you are at a velocity starting at 297

-32(t-3) +297 is fine, but it may help to distribute and combine the constants just to be safe

Answer 13

-32t +96+297 antiderives to what?

Answer 14

the distribution prolly isnt all that important

Answer 15

can I just add the 96+297? haha

Answer 16

you can yes

Answer 17

-16t^2+393t

Answer 18

right, or

-32(t-3)^2/2 + 297(t-3)+ the height it was at when this leg started

Answer 19

or maybe that takes care of the height .... never tried to not distribute first

Answer 20

mm well I guess it does since it's right xD o.o

Answer 21

wohoo

Answer 22

Mhm. Just the last one needs some work haha

Answer 23

when t>23

Answer 24

if something doesnt change, what do we call it?

Answer 25

... it remains the same, itis constant

Answer 26

what is the velocity at t=23?

Answer 27

I just plug that into the original velocity one? (the first one?)

Answer 28

of course not, the start of the last leg is determined by the leg before it ...

Answer 29

Wait that's before it slows down =.= sorry. Just a sec.

Answer 30

-23?

Answer 31

oh yay that's it. woop.
So now for position I have the first two done.

Answer 32

you should be able to do all of them now. antiderivatives of linear stuff is rather elementary and youve already done 2 of them

Answer 33

I always am messing up when I get to the constant part. Like trying to figure that out.

Answer 34

like 11t^3 I got because it is simply the anti derivative of the first one of velocity.

Answer 35

and when t=0,our starting position is .... well 0 isnt it?

Answer 36

Yep yep

Answer 37

we get our starting position from the last position we were at which is determined by the position equation above it

Answer 38

so the second we would get from..3?

Answer 39

plugging 3 into the first one to get the constant for the second one?

Answer 40

at 3 seconds we are starting at a position of 11(3)^3

Answer 41

yes

Answer 42

the third one's constant we would get from plugging 18 into the second?

Answer 43

yep

Answer 44

hmm that comes out to 1152. Is that correct? it seems large xD

Answer 45

16(t-3)^2-183(t-3)+1152
is incorrect.

Answer 46

because of the change of the time that it's falling at perhaps.

Answer 47

what we got for velocities are:

v1= 33t^2

v2= -32t +393

v3= 32t - 759

v4= -23

right?

Answer 48

yeah.

Answer 49

was position number 2 correct?

-16(t-3) +297(t-3) ?

Answer 50

ugh, forgot a ^2

Answer 51

position 2: -16(t-3)^2+297(t-3)+297 was correct

Answer 52

11*3^3 = 297 so that was our constant

Answer 53

what would you say is position 3?

Answer 54

uh well if we plug 18 into the second velocity we get -183

Answer 55

position, we want position not velocity

Answer 56

16(t-3)^2-183(t-3)+1152
for the third position but it's wrong. I Got the 1152 from plugging 18 into the second position.

Answer 57

-16(t-3)^2+297(t-3)+297, at t=18 is 1152 to start the next leg with

Answer 58

t-18 not t-3

Answer 59

oh gotcha. o.o Yeah it's right now then haha

Answer 60

and the last one?

Answer 61

637?

Answer 62

for the constant

Answer 63

637 for constant good

Answer 64

__(t-23)+637
I'm unsure what the first number is

Answer 65

-23 of course

Answer 66

v4= -23

p4 = -23(t-23)+ 637

Answer 67

for a constant: k

the antiderivative is: kt + c

Answer 68

oh yes yes. ok! so velocity and position are all correct now

Answer 69

i would hope so :)

Answer 70

haha yay!!
Now part b! xP

Answer 71

so max height would be when the first velocity = 0?

Answer 72

good luck with it ... ive got work in the morning so im heading out

Answer 73

Haha thanks so much for all your help :)

Answer 74

max height is when velocity = 0

test each velocity for 0 and see if t is in the interval

Answer 75

in some cases, it may be a min or a max so knowing how it looks graphically might be useful

Answer 76

okies!

Answer 77

good luck

Answer 78

thanks :)