# kinematics of a point in rectilinear motion

by xzibition8612
Tags: kinematics, motion, point, rectilinear
 P: 130 1. The problem statement, all variables and given/known data 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? 2. Relevant equations x''=a x'=at+c1 x=(a/2)t^2+(c1)(t)+c2 3. 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.
P: 602
 Quote by xzibition8612 Now plug in t=5 s into the first set of equations and get: 75+(5)(v0)=x1 30+v0=v1
How did you get the highlighted equation ?
 P: 130 plug t=5 into x'=6t+v0 Then knowing the fact that at t=5 the velocity is described by x'=-6t^2+v1, I set t=0 at this instant and thus 30+v0 = v1
P: 602

## kinematics of a point in rectilinear motion

you are using same constants on integration in different equations. the motion from t=5 to t=7
should have separate constants of integration from the motion t=7 onwards. thats why you are
mixing equations and got the wrong answer
 P: 130 But according to the problem, the motion at t=7s follows the same equation as the motion at t=5s (x''=-12), so shouldn't it be the same equations? Can you show me the correct equation? Because from the way I'm reading the problem my way to me is correct, when in fact it's not. I can't see it.

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