Question

Help please on two questions?

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(2t). Find the acceleration of the particle when the velocity is first zero.

The position of a particle on the x-axis at time t, t > 0, is s(t) = ln(t) with t measured in seconds and s(t) measured in feet. What is the average velocity of the particle for 1 ≤ t ≤ e?

Answer 1

take the second derivative to get the equation for the acceleration

Answer 2

That's what I'm stuck on, not exactly how the second derivative would look with ln and e

Answer 3

@satellite73

Answer 4

there is no \(e\) in it

Answer 5

\[s(t)=t\ln(2t)\] the derivative requires the the product rule

Answer 6

I got ln(2t) + 1 as the first derivative and 1/t as the second derivative??

Answer 7

Also my answer choices contain e so I was a little confused

Answer 8

ok first derivative is what you have
that is the velocity
you have to set that equal to zero and solve

Answer 9

How do I get rid of the ln?

Answer 10

you mean how to you solve \[\ln(2t)+1=0\]?

Answer 11

Yes

Answer 12

subtract 1 first then write in equivalent exponential form

Answer 13

i can show you if that is unclear

Answer 14

I understand the ln(2x) = 1 part but after that was unclear

Answer 15

-1 sorry

Answer 16

actually it should be \[\ln(2t)=-1\]

Answer 17

then \[2t=e^{-1}\]

Answer 18

so \[t=\frac{1}{2e}\]

Answer 19

OH! Okay that makes sense!

Answer 20

now your answer should be easy since as you wrote the second derivative is \(\frac{1}{t}\) and the reciprocal of \(\frac{1}{2e}\) is \(2e\)

Answer 21

Okay I understand now, thank you! and for the second one would I only need the first derivative?

Answer 22

no i don't think so

Answer 23

average is just the slope of the tangent line, unless i am misreading the question

Answer 24

So I just find the slope between that interval?

Answer 25

yes change in y over change is x
or change in s over change in t, whatever

Answer 26

Okay, thank you so much!!!