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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.