Question

help @vocaloid

Answer 1

need part b

Answer 2

velocity is the integral of acceleration
position function is the integral of the velocity function

Answer 3

I;m confused as to what to take the integral of for position because each thing I've tried I got the answer wrong

Answer 4

@Vocaloid

Answer 5

I have 7 more questions to do after this and they have to be done today been working on it since Tuesday and still have another lesson to do

Answer 6

acceleration = 6
velocity is the integral of 6 ---> v = 6t + C
the object starts at rest, so v = 0 at t = 0, making C = 0

position is the integral of velocity ---> p = 3t^2 + C
initial position is 5
5 = 3(0)^2 + C
C = 5

Answer 7

one more question of this part and then 4 more questions after this

Answer 8

wrong one

Answer 9

have part e need d

Answer 10

I'm nost sure what I am integrating

Answer 11

distance is the integral of the absolute value of the velocity function

Answer 12

If I use theoriginal equation I get 6 and the answer is 11.83?

Answer 13

@Vocaloid do I use 10t-4t^2
or 5t^-4/3t^3 or something else?

Answer 14

it is asking for the distance the car drove

the velocity of the car is 10t-4t^2

Answer 15

I plugged in 3 and got 6x?

Answer 16

x is not a variable in this question.

Answer 17

because I did absolute value

Answer 18

okay 6t

Answer 19

but still?

Answer 20

\[\int\limits_{0}^{3}\left| 10t-4t^{2} \right|dt\] you had it right the first time

Answer 21

omg thank you forgot the ranges and I'll send my last 4 in a sec

Answer 22

c-f

Answer 23

I know for c the disc dromula is vol=pi integral a-b r^2 dx

Answer 24

confused about the rotation part

Answer 25

I don't really remember how to do this

Answer 26

none of the 4?

Answer 27

maybe try khanacademy or something