Question

question about displacement!

Answer 1

36 right??

Answer 2

@Loser66 @DanJS

Answer 3

is it positive 36? or negative 36?

Answer 4

positive

Answer 5

oh wait!

Answer 6

displacement would be negative 36

Answer 7

you sure?

Answer 8

no...

Answer 9

how did you get +36 or -36 in the first place?

Answer 10

i thought displacement would be negative, but im wrong

Answer 11

no I didn't say you were right or wrong

Answer 12

ill upload my work, one sec

Answer 13

alright

Answer 14

ok its not working

Answer 15

so -6^2 is 36

Answer 16

the "attach file" button isn't working?

Answer 17

it attaches, it uploads, i refresh the page and it isnt there

Answer 18

hmm odd

Answer 19

one sec

Answer 20

You use the definite integral to find the exact net signed area under the curve. This area is equal to the displacement

\[\Large \int_{0}^{6}(v(t))dt\]


\[\Large \int_{0}^{6}(-t^2 + 6)dt\]


\[\Large \left. -\frac{t^3}{3}+6t+C\right]_{0}^{6}\]


\[\Large \left[-\frac{6^3}{3}+6(6)+C\right] - \left[-\frac{0^3}{3}+6(0)+C\right]\]


\[\Large -36\]

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So, \[\Large \int_{0}^{6}(-t^2 + 6)dt = -36\]

Answer 21

The answer is indeed -36, which means the object is displaced 36 feet in the negative direction. However, I think you used the wrong steps to get your answer.

Answer 22

thank you for the explanation! im still going over it, my teacher didnt explain this to me because i wasnt in school

Answer 23

hopefully this is calculus class? or is this some other class?