# Determine the acceleration after 6 seconds.

## Homework Statement

Determine the acceleration after 6 seconds.

## Homework Equations

s(t) = sqrt12t - 8 where s is in metres and t is in seconds.

## The Attempt at a Solution

v=s'(t)=6/sqrt12t-8

this is where i don't understand...they got a positive value for 36 in the book and i got a negative value...here is my work below that got me to a negative value...what did i do wrong?

a=v'(t)=0*sqrt12t-8 - 6/sqrt12t-8*6/(sqrt12t-8)^2
= - 36/sqrt12t-8/(sqrt12t-8)^2
= - 36/(12t-8)^3

Hard to tell what your original equation is..

$\sqrt{12t-8}$
or
$\sqrt{12t}-8$

??

It's kind of hard for me to see what any of your work is too.

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
Do you mean s = sqrt(12t - 8) or s = sqrt(12t) - 8? Use brackets, or else typeset your material in TeX.

If you mean s = sqrt(12t - 8), then v = ds/dt = 6/sqrt(12t - 8). If you mean s = sqrt(12t) - 8,
then v = (1/2)sqrt(12/t).

RGV

From the work done already, it looks to be the first option.

Could you explicitly state the entire question, also? Often times with acceleration/velocity problems the manner in which the problem is set up/interpreted can account for the sign differences that using strictly mathematics doesn't show.

And as mentioned, your work is very confusing. But, from what I can see, are you trying to use the quotient rule in getting the acceleration? What you need to use is the chain rule (which I am assuming you did for the correct velocity function) after moving the square root expression in the denominator up. Remember your power/exponent rules.

From the work done already, it looks to be the first option.

Could you explicitly state the entire question, also? Often times with acceleration/velocity problems the manner in which the problem is set up/interpreted can account for the sign differences that using strictly mathematics doesn't show.

And as mentioned, your work is very confusing. But, from what I can see, are you trying to use the quotient rule in getting the acceleration? What you need to use is the chain rule (which I am assuming you did for the correct velocity function) after moving the square root expression in the denominator up. Remember your power/exponent rules.

Sorry, the full question is: Given the position function s(t) = sqrt12t-8 where s is in metres and t is in seconds. Determine the acceleration after 6 seconds.

First i did this:

v=s'(t)=1/2(12t-8)^-1/2(12)

therefore v(t)=6/sqrt12t-8

a = v'(t)=d(6)/dt*sqrt12t-8 - d(sqrt12t-8)/dt*6/(sqrt12t - 8)^2

Hrm, I worked the problem and I also get a negative value. I also checked the graph and by eye it looks like it should be just barely negative.

Are you sure your initial function isn't a velocity function (and thus you'only derive once, that gives a positive value.)

Ya i am sure, but it could be that the answer in this book is wrong? The initial function is(according to the book) a position function.

I could have made an error, but the graph points to a slightly falling acceleration due to it's concavity.

a(6) = 0.07m/s^2

Then in my opinion, the book either has a sign error or your function was copied incorrectly.

Ray Vickson
Homework Helper
Dearly Missed
No. A correct statement would be: therefore, v(t) = 6/sqrt(12t - 8). I don't understand what you have against writing expressions clearly, using brackets. It really does not require extra work.

RGV

No. A correct statement would be: therefore, v(t) = 6/sqrt(12t - 8). I don't understand what you have against writing expressions clearly, using brackets. It really does not require extra work.

RGV

ya it does on my computer...nothing "against" writing expressions clearly...that is silly.

Ray Vickson
Homework Helper
Dearly Missed
I don't believe you. Before, you wrote v=s'(t)=1/2(12t-8)^-1/2(12) and that has lots of brackets.

RGV

i don't care buddy...you need a holiday.

Right... about the mathematics, can anyone else confirm that the book's solution does not match the problem we have been given?

The answer in the book is wrong. It showing a positive acceleration when it should be a negative value. The book made a sign error.

$$a(t) = v'(t) = \frac{-36}{(12t-8)^{\frac{3}{2}}}$$

For t = 6 sec

$$a(6) = v'(6) \, \approx \, -0.07 m/sec^{2}$$

Yes, the book must be wrong if we're all getting matching answers.

No surprise there. i come across a lot of mistakes in textbooks.