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?
The Attempt at a Solution
Now plug in t=5 s into the first set of equations and get:
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.