kinematics of a point in rectilinear motion


by xzibition8612
Tags: kinematics, motion, point, rectilinear
xzibition8612
xzibition8612 is offline
#1
Mar8-12, 04:17 PM
P: 135
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.
Phys.Org News Partner Science news on Phys.org
Simplicity is key to co-operative robots
Chemical vapor deposition used to grow atomic layer materials on top of each other
Earliest ancestor of land herbivores discovered
issacnewton
issacnewton is offline
#2
Mar9-12, 12:46 AM
P: 606
Quote Quote by xzibition8612 View Post
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 ?
xzibition8612
xzibition8612 is offline
#3
Mar9-12, 10:05 AM
P: 135
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

issacnewton
issacnewton is offline
#4
Mar9-12, 11:02 AM
P: 606

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
xzibition8612
xzibition8612 is offline
#5
Mar13-12, 03:57 PM
P: 135
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.


Register to reply

Related Discussions
Rectilinear Motion Introductory Physics Homework 2
Kinematics: Rectilinear Motion (calculus) Introductory Physics Homework 5
kinematics of rectilinear motion - multiple displacements Introductory Physics Homework 1
Rectilinear Kinematics Help Engineering, Comp Sci, & Technology Homework 3
Rectilinear motion Calculus & Beyond Homework 1