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)

Question

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)

anonymous · Answer

instead of x, it is t?

anonymous · Answer

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

phi · Answer

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

anonymous · Answer

cube root of a cube root? or what?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

3+1=4 lol

phi · Answer

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

anonymous · Answer

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

anonymous · Answer

is the answer -0.333?

phi · Answer

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

anonymous · Answer

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

phi · Answer

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

anonymous · Answer

Yup

anonymous · Answer

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

anonymous · Answer

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

phi · Answer

with a minus sign

anonymous · Answer

with a minus sign lol, make it negative

anonymous · Answer

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

anonymous · Answer

Now you can plug in 3