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)

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- anonymous

instead of x, it is t?

- anonymous

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

- phi

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

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## More answers

- anonymous

cube root of a cube root? or what?

- anonymous

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

- anonymous

\[3\sqrt[3]{4}\]

- anonymous

3+1=4 lol

- phi

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

- anonymous

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

- anonymous

is the answer -0.333?

- phi

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

- anonymous

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

- phi

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

- anonymous

Yup

- anonymous

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

- anonymous

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

- phi

with a minus sign

- anonymous

with a minus sign lol, make it negative

- anonymous

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

- anonymous

Now you can plug in 3

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