Ap Calculus

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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

  • Ap Calculus -

    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

  • Ap Calculus -

    I get an answer of 13 but it's not one of the choices

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