Mathematics, 30.11.2019 00:31 mustafajibawi1
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.
1 ft/sec2
two thirds ft/sec2
negative 1 over 108 ft/sec2
none of these
Answers: 3
![\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies \boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\ -------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}](/tpl/images/0396/5217/efd01.png)
  
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The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1...
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