## anonymous 4 years ago What is the acceleration at time t=3 of a particle whose position at time t is given by s(t)=3sqrt(x+1)^(1/3) Note: The square root is the third root and it is being multiplied by another 3 that's the 3 in front.. so, it's 3 * third root(x+1)

1. anonymous

instead of x, it is t?

2. anonymous

$s(3)=3(\sqrt[3]{x+1})^{1/3}=3\sqrt[9]{4}$

3. phi

$s(t)= 3(t+1)^{\frac{1}{3}}$

4. anonymous

cube root of a cube root? or what?

5. anonymous

it looks like rickjbr but without the 1/3... sorry

6. anonymous

$3\sqrt[3]{4}$

7. anonymous

3+1=4 lol

8. phi

replace the cube root with the equivalent to the 1/3 power. take the 2nd derivative of s(t) to find the acceleration

9. anonymous

oh that is the position...i'm retarded

10. anonymous

is the answer -0.333?

11. phi

you mean (t+1) not 4. plug in t=3 after the 2nd derivative

12. anonymous

whoops, i'm not thinking clearly, phi is right.

13. phi

you are in a rush to get the answer. Must be close to quitting time.

14. anonymous

Yup

15. anonymous

First derivative is (t+1)^(-2/3)

16. anonymous

2nd would be $\frac{2(t+1)^{-5/3}}{3}$

17. phi

with a minus sign

18. anonymous

with a minus sign lol, make it negative

19. anonymous

$\frac{-2}{3(t+1)^{5/3}}$

20. anonymous

Now you can plug in 3