The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 2 seconds.

Question

anonymous · Answer

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

anonymous · Answer

@ganeshie8

anonymous · Answer

hi i heard you were really good at math :) @ganeshie8 can you please help me with this equation?

anonymous · Answer

@Directrix

welshfella · Answer

the acceleration is given by  the second derivative of s(t)

welshfella · Answer

are you familiar with differentiation in calculus?

anonymous · Answer

yes :)

anonymous · Answer

finding the derivative, so is that the first step

anonymous · Answer

@Destinymasha

welshfella · Answer

yes find the derivative of (t^3 + 1)^ (1/2)

anonymous · Answer

ok 3t^2 / 2 sqrt (t^3+1)

welshfella · Answer

right 
now differentiate again

anonymous · Answer

then do i just plug in 2 into t?

anonymous · Answer

oh ok , so differentiate again then plug 2 into t? :)

welshfella · Answer

no - that is the velocity - to find the acceleration you have to differentiate again

welshfella · Answer

yes

welshfella · Answer

a bit messy ...

anonymous · Answer

kk so i got: 3t(t^3+4)/[4(t^3+1)^3/2]

anonymous · Answer

so the final answer is: 2/3 ? :)

anonymous · Answer

Which of the following expressions is the definition of the derivative of f(x) = cos(x) at the point (2, cos(2))?

anonymous · Answer

i have another question if u dont mind me asking? :)

anonymous · Answer

@e.mccormick

welshfella · Answer

i got 1.89 for the acceleration

welshfella · Answer

I made  t" = [12t(t^3 + 1) - 3t^2] / (4(t^3 + 1)^(3/2)