Question

guys please help me. I give medals

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

@jim_thompson5910

Answer 2

To find the time when the particle is first time at the origin, set x(t) = 0.
From the nature of cos() function, we know x(t) becomes zero for the first time when t√ equals π2. This gives the time instant.

To find the velocity at this time, differentiate x(t) with respect to t to get v(t), and plug in the value of time in the expression for v(t).

Answer 3

@triciaal

Answer 4

@jim_thompson5910

Answer 5

`Find the acceleration`

v(t) = s ' (t)
a(t) = v ' (t) = s '' (t)

so you'll have to find the second derivative of s(t). Equivalently, the first derivative of v(t)

Answer 6

if s(t) = t*ln(3t), then what is s ' (t) ?

Answer 7

so the acceleration is the second derivative right?

Answer 8

yes of the position function

Answer 9

ok, give me minute

Answer 10

i think it's wrong, but I got 1+ln(3x) for the first derivative

Answer 11

pretty!
Find the acceleration of the particle when the velocity is first zero
so you have
v=1+ln(3x)
you want to find x such that v is 0

Answer 12

then you find v'

Answer 13

s(t) = t*ln(3t)

s ' (t) = d/dt [ t*ln(3t) ]

s ' (t) = d/dt [ t ]*ln(3t) + t* d/dt [ ln(3t) ]

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

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

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

so you have it correct @Joseluess

Answer 14

ok, so the secod derivative is 1/x

Answer 15

1/t, yes

Answer 16

okay so now do I plug zero ?

Answer 17

well it says to evaluate the acceleration when the velocity is 0
that is why I asked you to find x such that v is 0

Answer 18

v=1+ln(3x)
0=1+ln(3x)


note: I don't know why it says when the velocity is first 0
because it is only ever 0 once

Answer 19

but I have to solve for the acceleration, which is the second derivative.

Answer 20

yes I know

Answer 21

but you need to evaluate the acceleration at the value that makes the velocity 0

Answer 22

so do i solve for x when velocity is zero?

Answer 23

yes solve 0=1+ln(3x)

Answer 24

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

as freckles is saying, you plug v = 0 into v = 1+ln(3t) and solve for t

then you take that t value and plug it into a(t) = 1/t

Answer 25

ok, I got stuck I completely forgot how to solve from this part

Answer 26

do I ln both sided to cancel ln ?

Answer 27

\[\log_b(a)=x \text{ has equivalent exponential form } b^x=a \\ \text{ also note that } \ln(x)=\log_e(x)\]

Answer 28

1/3e

Answer 29

yes, \[\Large t = \frac{1}{3e}\] now plug that into a(t)

Answer 30

3e

Answer 31

yep, a(t) = 3e is your acceleration when t = 1/(3e)

Answer 32

okay, thank you so much..... I'll give it to you the medal, but @freckles I'll make another question and ill give it to you

Answer 33

no I don't really care
I just wanted to help with a calculus problem even though @jim_thompson5910 was already doing an awesome job
because lets face it calculus is most fun