A. Find the position when the acceleration is equal to -28.
B. Find the acceleration when t=3
2007-10-30 12:48:51 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x' =4t^3+12t^2-8t
x" = 12t^2 +24t-8
a = x"(3) = 12*9 +24(3)-8 = 172 units/sec^2
12t^2 +24t -8 = 28
12t^2 +24t-36 = 0
t^2 +2t - 3 = 0
(t+3)(t-1) = 0
t = 1
x(1) = 4+4-8 = 0 units
2007-10-30 12:58:34 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
Take the second derivative to get the acceleration function. You can plug in T=3 into that function to answer B.
Furthermore, when acceleration is equal to -28 you set that equal to your second derivative of the position function and you solve for T. Then you plug in T into the position function and you solve it for x.
2007-10-30 12:53:21 · answer #2 · answered by Mike A 2 · 0⤊ 0⤋
Go to the library
2007-10-30 12:54:22 · answer #3 · answered by Diana$@$ 6 · 0⤊ 1⤋
-4t^5
2007-10-30 12:52:59 · answer #4 · answered by ♥♥ 5 · 0⤊ 1⤋