Ap Calculus
posted by Anonymous on .
Please help.
A particle moves in a stright line and its acceleration is given by a(t)=6t+4. its initial position is s(0)=9 and its position when t=1 is s(1)=6. find the velocity of the particle when t=2.
A. v(2)=14
B. v(2)=7
C. v(2)=4
D. v(2)=9
E. v(2)=0

a = 6 t+4
then
v = 3 t^2 + 4 t + c
then
s = t^3 + 2 t^2 + c t + k
now find the constants c and k from the given points
s(0) = 9
9 = 0^3 + 2 0^3 + c 0 + k
so
k = 9 and we so far have
s = t^3 + 2 t^2 + c t + 9
now
s(1) = 6
6 = 1^3 + 2 (1)^2 + c (1) + 9
6 = 1 + 2 + c + 9
c = 6
so in the end
s = t^3 + 2 t^2  6 t + 9
I guess you can put in t = 2 
I get an answer of 13 but it's not one of the choices