The position equation for the movement of a particle is given by s = (t^2 - 1)^3 when s is measured in feet and t is measured in seconds. Find the acceleration at two seconds.
I got 90 units/sec^2....
Is that right? If not, can someone explain it to me?
2007-04-26 11:22:42 · 2 answers · asked by b_ascha 2 in Science & Mathematics ➔ Mathematics
acceleration (change of rate) is the second derivative of an equation- kinda like the first derivative is the slope of a function, the second derivative is the change in slope...
s = t^6 - 3(t^4) + 3(t^2) - 1
first derivative = 6(t^5) - 12(t^3) + 6t
second derivative = 30(t^4) - 36(t^2) + 6
at 2 seconds (t = 2) acceleration = 30*16 - 36*4 + 6 ft/sec^2
acceleration = 342 ft/(sec^2)... hopefully that helped!
2007-04-26 11:33:56 · answer #1 · answered by SkiBum 4 · 1⤊ 0⤋
To determine acceleration, you must compute the second derivative of s: s``=[6(t^2-1)2*t]`=6(t^2-1)^2+12(t^2-1)*2t^2
so a(2)=6*9+12*3*2*4=342.
2007-04-26 11:34:00 · answer #2 · answered by bruinfan 7 · 1⤊ 0⤋