Posted by **Anonymous** on Sunday, January 27, 2013 at 2:32pm.

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 -
**Damon**, Sunday, January 27, 2013 at 2:50pm
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 -
**Anonymous**, Sunday, January 27, 2013 at 4:32pm
I get an answer of 13 but it's not one of the choices

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