Question

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(3t). Find the acceleration of the particle when the velocity is first zero

Answer 1

are you able to take the derivative?

Answer 2

yes

Answer 3

what do you get when you take the derivative

Answer 4

ln(t)+ln(3)+1

Answer 5

which is the same as ln(3t) + 1, good

Answer 6

set s ' (t) equal to zero and solve for t


s ' (t) = 0

ln(3t) + 1 = 0

t = ??

Answer 7

t=-.9102

Answer 8

I'm getting a positive value

Answer 9

Try again

Answer 10

I got the same thing. I subtracted the -1 to the other side then divided by ln3 that gave me what i have

Answer 11

ln(3t) + 1 = 0

ln(3t) = -1

3t = e^(-1)

t = ??

Answer 12

.1226

Answer 13

good

Answer 14

this is when the velocity is zero

Answer 15

now you need to find s '' (t), the acceleration function

Answer 16

how do i do that?

Answer 17

do i take the derivative of the derivative?

Answer 18

yes, you would take the derivative of s ' (t)

Answer 19

would it be ln3/x?

Answer 20

I'm getting s '' (t) = 1/t

Answer 21

be sure to simplify fully (and use the chain rule)

Answer 22

im confused. what did you do exactly?

Answer 23

s ' (t) = ln(3t) + 1

s '' (t) = 1/(3t)*3

s '' (t) = 1/t

Answer 24

I derived ln(3t) to get 1/(3t) and then you have to multiply by the derivative of the inside stuff (the derivative of 3t, which is 3)

so using the chain rule

if y = ln(3t), then dy/dt = 1/(3t)*3 = 1/t

Answer 25

Okay so then you have 1/t do I just fill in what i got for t earlier into that formula?

Answer 26

"Find the acceleration of the particle when the velocity is first zero"

Answer 27

so you plug t = 0.1226 into s '' (t) to get the acceleration at t = 0.1226

Answer 28

so 8.1548?

Answer 29

correct

Answer 30

thats not it is it?

Answer 31

it gave me answer choices and they are as followed

a. 3e^2
b. e
c. 3e
d. none of the above
im guessing that means its d

Answer 32

hmm one sec

Answer 33

let's go back to ln(3t) + 1 = 0

Answer 34

ln(3t) + 1 = 0

ln(3t) = -1

3t = e^(-1)

3t = 1/e

t = 1/(3e)

so at exactly t = 1/(3e), the particle has velocity 0

Answer 35

now plug t = 1/(3e) into the acceleration function s '' (t)

s '' (t) = 1/t

s '' (1/(3e)) = 1/[ 1/(3e) ]

s '' (1/(3e)) = 3e

so the acceleration is exactly 3e

Answer 36

Okay then that makes more sense thank you!

Answer 37

you're welcome