Question

Hello, this is my first time using OpenStudy aside from browsing through how to do certain problems. Here's one I'm having trouble on:

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.

I tried getting the second derivative and plugging in zero, but that doesn't seem to be right. Could somebody help?

Answer 1

The acceleration is the second derivative

Answer 2

\(\color{#000000 }{ \displaystyle s(t)=t\ln(3t) }\)
this is your position function, correct?

Answer 3

That is correct.

Answer 4

In general, the derivative of the function is the function's slope, and this derivative will tell us how fast (or how "steep") the function is going at every point.

And so, the derivative (the slope) of the position tells us how fast this position will move. So, wouldn't you agree that the derivative of the position, is the velocity?

Answer 5

Yes, and the derivative of velocity is the acceleration.

Answer 6

can you differentiate the s'(t)?
(to find the general expression for velocity of the particle)?

Answer 7

I mean find the s'(t) (not differentiate s'(t))....

Answer 8

\(\color{#000000 }{ \displaystyle s(t)=t\ln(3t) }\)

Answer 9

use the product rule to differentiate this

Answer 10

ln(3t) + 1, correct?

Answer 11

\(\color{#000000 }{ \displaystyle s(t)=t\ln(3t) }\)
\(\color{#000000 }{ \displaystyle s(t)=t\ln(t)+t\ln(3) }\)

\(\color{#000000 }{ \displaystyle s'(t)=1\cdot\ln(t)+t\cdot \frac{1}{t}+\ln(3) }\)
\(\color{#000000 }{ \displaystyle s'(t)=\ln(t)+1+\ln(3) }\)

yes, your derivative is corret

Answer 12

Now, you are asked to find the acceleration at the points where the particle's velocity is zero.

Answer 13

Do I plug in zero, then take the derivative of that?

Answer 14

Wait, set it to zero rather.

Answer 15

shouldn't the domain be t>0? as t ->0+, ln(t) -> - infinity

Answer 16

I don't think so. That's the domain I was given.

Answer 17

so what's the position at t = 0? doesn't make sense.

Answer 18

sorry, I don't mean to distract...

Answer 19

@solomonzelman Any insight you'd be willing to give? I'm rather confused.

Answer 20

Of course aside from guiding me right up to the point where I was initially confused--my initial steps were shown to be correct.

Answer 21

I had technological difficulties... apologize

Answer 22

pgpilot326, it is asking for the points where velocity is 0, not s'(0)

Answer 23

Don't sweat it! I really appreciate your guidance.

Answer 24

\(\color{#000000 }{ \displaystyle s(t)=t\ln(3t) }\)
Find the acceleration of the particle when the velocity is first zero.

\(\color{#000000 }{ \displaystyle s'(t)=1\cdot \ln(3t) +t\times \frac{1}{3t}\times 3 = \ln(3t)+1 }\)

You need to know the point(s) where velocity is =0.

\(\color{#000000 }{ \displaystyle 0=1\cdot \ln(3t) +t\times \frac{1}{3t}\times 3 = \ln(3t)+1 }\)

Answer 25

\(\color{#000000 }{ \displaystyle 0= \ln(3t)+1 }\)

Answer 26

\(\color{#000000 }{ \displaystyle -1= \ln(3t) }\)
\(\color{#000000 }{ \displaystyle -1= \log_e(3t) }\)
\(\color{#000000 }{ \displaystyle e^{-1}= 3t }\)

go on...

Answer 27

the solution to that (to s'(t)=0), is going to give you the point where the velocity of the particle is zero (say that you find that \(x=\beta\). (Just making it up)

Keep that in mind, because you want to find the acceleration at the point \(x=\beta\) \( (\)since the question asks for acceleration at point(s) where velocity is zero\( ) \), so you need to find the second derivative \(s''(t)\) which gives you the acceleration for \(s(t)\) at every point, and then to plug in \(s''(\beta)\) to find the acceleration at a point \(x=\beta\) \( (\) Which is the acceleration wanted - the acceleration of the particle at the point where the velocity is equal to zero\( )\).

This is verbal, and if you want I can give you an example similar to this.

Answer 28

Alright so I ended up with t = 1/3e ; however, that's not a choice. Forgive me. I completely understand the conceptual aspects you've pointed out, but this isn't a choice. Hm.

Answer 29

1/(3e) is just a point where velocity is 0.

Answer 30

yiou will need acceleration at this point, so find the second derivative of the function s(t), and evaluate it at t=1/(3e)

Answer 31

So, you need f''(1/(3e))

Answer 32

Yes you're right!! Sorry haha I got it now!

Answer 33

Thanks so much. That was dumb of me.

Answer 34

No problem.... unlike many users you preform the work correctly, and I enjoy that! Ty and yw ;)

Answer 35

welcome @Jerobrien !!!

Answer 36

Similar example (just in case):

A particle is moving along the x-axis so that its position
at t ≥ 0 is given by s(t) = 2t•In(6t). Find;

\(\small\color{blue}{(a)}\) Acceleration of the particle when the velocity is =0.
\(\small\color{blue}{(b)}\) Acceleration of the particle when the velocity is =2.
\(\small\color{blue}{(c)}\) Prove that the particle's velocity always increases.

\(\color{#000000 }{ \displaystyle s(t)=2t\ln(6t) }\)

Velocity/Derivative
\(\color{#000000 }{ \displaystyle s'(t)=2\ln(6t)+2 }\)

Velocity is 0,
\(\color{#000000 }{ \displaystyle 0=2\ln(6t)+2 }\)
\(\color{#000000 }{ \displaystyle -2=\ln(6t) }\)
\(\color{#000000 }{ \displaystyle e^{-2}=6t }\)
\(\color{#000000 }{ \displaystyle t=1/(6e) }\)

Velocity is 2,
\(\color{#000000 }{ \displaystyle 2=2\ln(6t)+2 }\)
\(\color{#000000 }{ \displaystyle 0=\ln(6t) }\)
\(\color{#000000 }{ \displaystyle t=1/6}\)

Acceleration/2nd Derivative
\(\color{#000000 }{ \displaystyle s''(t)=2/t }\)

Acceleration at the point where velocity is 0,
\(\color{#000000 }{ \displaystyle s''(1/(6e))=2/(1/(6e)) }\)
\(\color{#000000 }{ \displaystyle s''(1/(6e))=12e }\) ANSWER \(\small\color{blue}{(a)}\)

Acceleration at the point where velocity is 2,
\(\color{#000000 }{ \displaystyle s''(1/6)=2/(1/6) }\)
\(\color{#000000 }{ \displaystyle s''(1/(6e))=12 }\) ANSWER \(\small\color{blue}{(b)}\)

The time is always positive, and for all positive t,
\(\color{#000000 }{ \displaystyle s''(t)=2/t }\) is going to be positive. And positive
acceleration, by definition is the increase in velocity.
((accel. is always pos. = veloc. always increases))
ANSWER \(\small\color{blue}{(c)}\)

Answer 37

Although velocity is increasing, the acceleration is slowing down;
\(\color{#000000 }{ \displaystyle \frac{ds }{dt} \left[\frac{2}{t}\right] =-\frac{2}{t^2}}\)

And the derivative of accelration is always negative, therefore the acceleration is decreasing.

Answer 38

@Solomonzelman thanks so much again! I went through that example of yours and I do understand it completely now. Also, thanks @irishboy123 :)

Answer 39

Anytime!
Enjoy!