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?
Can someone show me how to do this?

Question

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?
Can someone show me how to do this?

kaleidoscopicsink · Answer

@Loser66

kaleidoscopicsink · Answer

@mathmale

kaleidoscopicsink · Answer

@pooja195

aihberkhan · Answer

Okay.

Looking at the average velocity between $t=2e$ and $t=e$. You can see it here:

$$ \frac{ \ln(2e) - \ln (2e) }{ 2e - e }$$

kaleidoscopicsink · Answer

1
 e
 e − 1
1/e-1
I think it's d or b. But I'm not sure.

aihberkhan · Answer

Well... lets finish the work first.

Now let's simplify.

First let's simplify the denominator.

$2e - e$ will become $e(2 - 1)$. Then that will become, simply, $e$.

So we will have:

$$\frac{ \ln(2e) - \ln(2e) }{ e }$$

aihberkhan · Answer

Lastly, you have to us the log property for simplifying the numerator:

$$\ln(a) -\ln (b) = \ln (\frac{ a }{ b })$$

aihberkhan · Answer

Try solving.

kaleidoscopicsink · Answer

1
 e
 e − 1
1/e-1 
These are the answer choices.
Give me a second to attempt solving.

aihberkhan · Answer

Okay take your time! :)

kaleidoscopicsink · Answer

if you had ln(a/b) that would go to ln(2e/2e) so ln 1 which is 1/x therefore being d?

aihberkhan · Answer

Great Job! :) I would say that as well! :)

aihberkhan · Answer

Hope this helped! Have a great day! :) 

A medal would also be appreciated! Just click best response next to my name! 

Also, a fan would be appreciated as well! Just hover over my icon and click "Become A Fan". This will allow you to see every time I am online! :) 

If you see that I am online and need help with a question, just tag me in your question! 

@KaleidoscopicsInk

kaleidoscopicsink · Answer

Thank you so much for your help! I fanned you and will take you up on that offer most likely. cx Have a great day.