The position of a particle in rectilinear motion is given by s(t) = tln(1/t) for t>0, where t is in seconds.

Question

A. Find the time (t>0) at which the velocity of the particle is 0. (I got t=1/e sec.)

B. Find the particle's acceleration at that time as well. You must find exact values - don't approximate with a calculator. (I got -e)

Please check my answers and units for each answer. Also, are my answers fine as they are, or do I need to convert to decimals(in the question it says to not approximate with a calculator)?

Answer 1

first, note that s(t) = t ln(1/t) = -t lnt
s' = -lnt - 1
s'=0 when lnt = -1, or t = 1/e

s(1/e) = 1/e * 1 = 1/e

Better review logs some more. And always do a sanity check on answers.
At least you should have immediately recognized that logs of negative numbers are undefined!

Answer 2

So my answers are correct?

Answer 3

not so.
You better redo your own calculations.
What do you think is the derivative?

and show me how the velocity is -e. Please.

But yes, I do retract my comment about logs of negative numbers. That was the supposed answer. My bad.

Answer 4

A. v(t)=s'(t)=-1+ln(1/t)
-1+ln(1/t)=0
ln(1/t)=1
t=1/e sec

B. a(t)=v'(t)=-1/t
Plug in t: -1/(1/e) = -e

Answer 5

Also, the acceleration at the time t=1/e sec is -e, not the velocity!

Answer 6

A. You will note that our answers for s' actually agree, since ln(1/t) = -ln(t)

and you are right again about the acceleration.

Answer 7

So, according to u now, @Miller is correct for both parts, right?

Answer 8

The position function of a particle in rectilinear motion is given by s(t) s(t) = t3 - 9t2 + 24t + 1 for t ≥ 0. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer
=============================

Reversing direction means changing of velocity sign.
the motion is rectilinear, along a straight lin? that is; so V can be either + or -

And, velocity is:

V=ds/dt=3t^2-18t+24

This is a parabola, crossing "t" axes twice - in two moments in time the particle reverses its direction.:
We have to find the nearest "future" time - the smalllest "zero" for parabola V.

Let's see when V turns to zero; it's happening when

V=0 i.e. 3t^2-18t+24=0 , divide by 3

t^2-6t+8=0 , solve the quadratic:

solutions are t=2 and t=4

It's happening in positions at s(2) and s(4)
You have to find s(2)=... and s(4)=...


The acceleration is, a=dV/dT=6t-18
Further,
a(2)=6*2-18=-6
a(4)=+6

Units: .... The problem doesn't provisde any units here: "s" can be measured in cm, m,mi, ...
and the time- in s, min, hr, ...
In all cases, V has dimention of [distance]/[time]
a - [distance]/[time^2]

This must help.
Check the arithmetics, and complete the computations ...

Good luck!

P.S. my understanding is, problem requires a single solution (... find the position). Then we must consider the expressons for t=2; unless particle starts its motin in the moment t>2.
Doble check problem conditions ... the only initial state dta mentioned is: t>0. no units, no initial moment provided.

Answer 9

the person signing in as "mathhelper", switch to a different name, I have used that name for some time now.
I would not question oobleck, 99.99% of the time he is correct

Answer 10

Ok, can you please tell the correct answer for @Miller's question?