Determine the acceleration after 6 seconds.

[1irishman]

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

 

[1MileCrash]

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.

 

[Ray Vickson]

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

 

[Ecthelion]

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.

 

[1irishman]

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.

 

[1irishman]

First i did this:

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

 

[1irishman]

therefore v(t)=6/sqrt12t-8

 

[1irishman]

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

 

[1MileCrash]

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.)

 

[1irishman]

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.

 

[1MileCrash]

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

Was the book's answer. 75?

 

[1irishman]

the books answer was:
a(6) = 0.07m/s^2

 

[1MileCrash]

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

 

[Ray Vickson]

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

 

[1irishman]

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]

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

 

[1irishman]

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

 

[1MileCrash]

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

 

[Skins]

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

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

 

[1irishman]

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

 

[Skins]

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