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anonymous
 one year ago
guys please help me. I give medals
A particle is moving along the xaxis 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
anonymous
 one year ago
guys please help me. I give medals A particle is moving along the xaxis 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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To 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).

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3`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
 one year ago
Best ResponseYou've already chosen the best response.3if s(t) = t*ln(3t), then what is s ' (t) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the acceleration is the second derivative right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yes of the position function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think it's wrong, but I got 1+ln(3x) for the first derivative

freckles
 one year ago
Best ResponseYou've already chosen the best response.1pretty! 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3s(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
 one year ago
Best ResponseYou've already chosen the best response.0ok, so the secod derivative is 1/x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so now do I plug zero ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well 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
 one year ago
Best ResponseYou've already chosen the best response.1v=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
 one year ago
Best ResponseYou've already chosen the best response.0but I have to solve for the acceleration, which is the second derivative.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but you need to evaluate the acceleration at the value that makes the velocity 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do i solve for x when velocity is zero?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3`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
 one year ago
Best ResponseYou've already chosen the best response.0ok, I got stuck I completely forgot how to solve from this part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440985593903:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do I ln both sided to cancel ln ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\log_b(a)=x \text{ has equivalent exponential form } b^x=a \\ \text{ also note that } \ln(x)=\log_e(x)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yes, \[\Large t = \frac{1}{3e}\] now plug that into a(t)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yep, a(t) = 3e is your acceleration when t = 1/(3e)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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
 one year ago
Best ResponseYou've already chosen the best response.1no 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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