anonymous
  • anonymous
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
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@jim_thompson5910
anonymous
  • anonymous
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).
anonymous
  • anonymous
@triciaal

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anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
`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)
jim_thompson5910
  • jim_thompson5910
if s(t) = t*ln(3t), then what is s ' (t) ?
anonymous
  • anonymous
so the acceleration is the second derivative right?
jim_thompson5910
  • jim_thompson5910
yes of the position function
anonymous
  • anonymous
ok, give me minute
anonymous
  • anonymous
i think it's wrong, but I got 1+ln(3x) for the first derivative
freckles
  • freckles
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
freckles
  • freckles
then you find v'
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
ok, so the secod derivative is 1/x
jim_thompson5910
  • jim_thompson5910
1/t, yes
anonymous
  • anonymous
okay so now do I plug zero ?
freckles
  • freckles
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
freckles
  • freckles
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
anonymous
  • anonymous
but I have to solve for the acceleration, which is the second derivative.
freckles
  • freckles
yes I know
freckles
  • freckles
but you need to evaluate the acceleration at the value that makes the velocity 0
anonymous
  • anonymous
so do i solve for x when velocity is zero?
freckles
  • freckles
yes solve 0=1+ln(3x)
jim_thompson5910
  • jim_thompson5910
`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
anonymous
  • anonymous
ok, I got stuck I completely forgot how to solve from this part
anonymous
  • anonymous
|dw:1440985593903:dw|
anonymous
  • anonymous
do I ln both sided to cancel ln ?
freckles
  • freckles
\[\log_b(a)=x \text{ has equivalent exponential form } b^x=a \\ \text{ also note that } \ln(x)=\log_e(x)\]
anonymous
  • anonymous
1/3e
jim_thompson5910
  • jim_thompson5910
yes, \[\Large t = \frac{1}{3e}\] now plug that into a(t)
anonymous
  • anonymous
3e
jim_thompson5910
  • jim_thompson5910
yep, a(t) = 3e is your acceleration when t = 1/(3e)
anonymous
  • anonymous
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
freckles
  • freckles
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

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