Question

What am i doing wrong??

Answer 1

I need help with the problems that are marked in the red "X"

Answer 2

...

Answer 3

for the part where asking when the particle is moving in the positive direction, is time 0 and 4 suppose to be included or not?

Answer 4

same with time 6

Answer 5

sorry man this is way above me @carlyleukhardt maybe she can help u

Answer 6

same ^

Answer 7

@ganeshie8

Answer 8

For the first part I know my answer is right, but I'm not sure if im suppose to include time 0s and 4s, and time 6s.

Answer 9

well the particle can't be at rest and also moving in the positive direction at the same time

Answer 10

but I did another question like this, and it said I had to include time 0s

Answer 11

ive always thought the same as u

Answer 12

i would have said it was moving right when [0,4) union (6,inf)

Answer 13

looking at the graph
acceleration is in blue
acceleration is the change in velocity with time
when is the velocity positive and the time positive (0<t< 4)

Answer 14

Freckles I see what you mean now. when you said the particle cannot be stationary and moving in the same positive direction. is that why you did not include 4s and 6s?. cuz the instantaneous are zero at those times, thus the particle is stationary?

Answer 15

yes I did not include x=4 or x=6
because the particle is at rest there therefore the particle cannot be moving right at x=4 or at x=6

Answer 16

ok, I need help with the last two parts as well

Answer 17

from time 0 to 4s isn't the particle slowing down?

Answer 18

wait is the particle slowing down from 0 to 4s, and then 5 - 6s? cuz the particle changes direction right?

Answer 19

I think we need to find the second derivative
and see where velocity and acceleration agree on sign (this is where the particle is speeding up)
and see where velocity and acceleration disagree on sign (this is where the particle is slowing down)
And yes I think you are right... [0,4) union (5,6) seems to have a disagreement in signs between the acceleration and velocity which means the particle is slowing down there

Answer 20

oh you already found second derivative

Answer 21

yea