Kinematics of a point in rectilinear motion

  • #1

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.
 

Answers and Replies

  • #2
911
18
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 ?
 
  • #3
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
 
  • #4
911
18
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
 
  • #5
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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