Calculus 2nd Derivative

1. Nov 3, 2007

Destrio

A particle moves along the x-axis, its position at time t is given by
x(t)= 5t/(10+7t^2)

where t is >= 0 and measured in seconds and x is in meters

The acceleration of the particle equals 0 at time t = ?

I took the 1st derivative and got
i got v(t ) = 5(10-7t^2) / (10+7t^2)^2

or i got v(t ) = 5(10-7t^2) * (10+7t^2)^-2

then i took the derivative of that and got:

a(t ) = 5[(10-7t^2)(-2)(10+7t^2)^(-3) *(14t) + (10+7t^2)^2 *(10-7t^2)]

so i tried factoring out (10-7t^2), to get one answer which i got as sqrt(10/7)

and then for the other answer i got:
(10+7t^2)^5 = 28t

which i simplified to:
2401t^8 + 13720t^6 + 29400t^4 + 28000t^2 - 28t + 10000 = 0

now I'm not sure if I'm on the right track, and if I am, I'm not sure how I can solve for t.

Any help is much appreciated, I've been working on this (my first problem) for about an hour and I have a midterm in a few days :s

thanks

2. Nov 3, 2007

Galileo

Just keep the powers of (10+7t^2)^2 in the denominator like you did when calculating v. The numerator is then easy to factorize.

3. Nov 3, 2007

Destrio

the numerator will still end up being 2401t^8 + 13720t^6 + 29400t^4 + 28000t^2 - 28t + 10000
if i make a common denomintor with the two terms
ahhh

4. Nov 4, 2007

Galileo

Don't write out the brackets, that's gonna be horrible. If you never write out the brackets in your calculations you'll end up with a form that's already quite nicely factorized.

If you apply to quotient rule to:
5(10-7t^2)/(10+7t^2)^2
without expanding, you get (10+7t^2)^4 in the denominator right? YOu'r numerator will not be too bad either. Just look for common terms.

5. Nov 4, 2007

Destrio

what do you mean don't write out the brackets?

applying the quotient rule i got:
5[(10+7t^2)^2 *(-14t) - (10-7t^2)(2)(10+7t^2)(14t)]/(10+7t^2)^4

so I can factor out 14t
and 10+7t^2

will my two terms be
10+7t^2
and
10-7t^2
?

thanks

6. Nov 4, 2007

Destrio

why did I get something different from when I used to product rule?
or were they equivalent but arranged differently?

7. Nov 4, 2007

Galileo

You made a small error when using the product rule. The last answer you gave using the qoutient rule is correct and it should be easy finding the zero's now.