Question

http://prntscr.com/9yozqg

Answer 1

What part don't you get?

Answer 2

how to start it @NoelGreco

Answer 3

What is the sum of the positive areas?

Answer 4

3Q-1

Answer 5

is that correct?

Answer 6

@Owlcoffee @zepdrix

Answer 7

@Zarkon

Answer 8

For the positive area?
Yah that sounds right :)

Total Distance Traveled = (Q-2) + (2Q+1) - (2P+3) - (P-4)

Answer 9

Displacement is measuring how far the object ended up in relation to it's initial position, so ummmm

Answer 10

I thought distance is when you add all of it up even if it has a negative area

Answer 11

Oh yes yes yes :)
I'm being silly, sorry bout that.

Answer 12

Total Distance Traveled = (Q-2) + (2Q+1) + (2P+3) + (P-4)
Displacement = (Q-2) + (2Q+1) - (2P+3) - (P-4)

Answer 13

total distance = 3Q+3P-2
DIsplacement = 3Q-3P
Is that what you got?

Answer 14

Mmmmm ya that sounds right so far :)

Answer 15

ok cool, what's the next step?

Answer 16

Find the difference in these values.
Difference means to subtract, ya?

Answer 17

yes, so 3Q+3P-2-3Q-3P?

Answer 18

3Q+3P-2-(3Q-3P)

Answer 19

oops my bad

Answer 20

6P-2

Answer 21

Yay good job \c:/

I think that's what they were looking for.

Answer 22

but it's giving me the velocity and i'm finding position stuff, am i not taking any integrals?

Answer 23

These values involving P an Q are `the result of integration`.
They represent the area under the curve for specific intervals (between the intercepts).

Answer 24

So they already did the work of integrating :)
They wanted you to look at it in a different way I guess.

Answer 25

oooh ok I understand, thank you!

Answer 26

np