Posted by Anonymous on Sunday, January 27, 2013 at 2:32pm.
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.
- Ap Calculus - Damon, Sunday, January 27, 2013 at 2:50pm
a = 6 t+4
v = 3 t^2 + 4 t + c
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
k = 9 and we so far have
s = t^3 + 2 t^2 + c t + 9
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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