anonymous
  • anonymous
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)
Mathematics
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anonymous
  • anonymous
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)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
instead of x, it is t?
anonymous
  • anonymous
\[s(3)=3(\sqrt[3]{x+1})^{1/3}=3\sqrt[9]{4}\]
phi
  • phi
\[s(t)= 3(t+1)^{\frac{1}{3}}\]

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anonymous
  • anonymous
cube root of a cube root? or what?
anonymous
  • anonymous
it looks like rickjbr but without the 1/3... sorry
anonymous
  • anonymous
\[3\sqrt[3]{4}\]
anonymous
  • anonymous
3+1=4 lol
phi
  • 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
  • anonymous
oh that is the position...i'm retarded
anonymous
  • anonymous
is the answer -0.333?
phi
  • phi
you mean (t+1) not 4. plug in t=3 after the 2nd derivative
anonymous
  • anonymous
whoops, i'm not thinking clearly, phi is right.
phi
  • phi
you are in a rush to get the answer. Must be close to quitting time.
anonymous
  • anonymous
Yup
anonymous
  • anonymous
First derivative is (t+1)^(-2/3)
anonymous
  • anonymous
2nd would be \[\frac{2(t+1)^{-5/3}}{3}\]
phi
  • phi
with a minus sign
anonymous
  • anonymous
with a minus sign lol, make it negative
anonymous
  • anonymous
\[\frac{-2}{3(t+1)^{5/3}}\]
anonymous
  • anonymous
Now you can plug in 3

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