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xzibition8612
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Homework Statement
A point Q in rectilinear motion passes through the origin at t=0, and from then until 5 seconds have passed, the acceleration of Q is 6 ft/s^2 to the right. Beginning at t = 5s, the acceleration of Q is 12t ft/s^2 to the left. If after 2 more seconds point Q is 13 feet to the right of the origin, what was the velocity of Q at t=0?
Homework Equations
x''=a
x'=at+c1
x=(a/2)t^2+(c1)(t)+c2
The Attempt at a Solution
At t=0:
x''=6
x'=6t+v0
x=3t^2+(v0)(t)
At t=5:
x''=-12t
x'=-6t^2+v1
x=-2t^3+(v1)(t)+x1
At t=7:
x''=-12t
x'=-6t^2+v1
13=-2t^3+(v1)(t)+x1
Now plug in t=5 s into the first set of equations and get:
75+(5)(v0)=x1
30+v0=v1
Plug these two equations into 13=-2t^3+(v1)(t)+x1
As for t in this equation, plug in 2 seconds because this new equation started at t=5 (and hence assume that to be the new starting point, thus 7-5=2).
I get v0 = -15 m/s
The book says the answer is +2. Obviously something went very wrong. Any help would be appreciated.