Register to reply

Kinematics of a point in rectilinear motion

by xzibition8612
Tags: kinematics, motion, point, rectilinear
Share this thread:
xzibition8612
#1
Mar8-12, 04:17 PM
P: 136
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
Wildfires and other burns play bigger role in climate change, professor finds
SR Labs research to expose BadUSB next week in Vegas
New study advances 'DNA revolution,' tells butterflies' evolutionary history
issacnewton
#2
Mar9-12, 12:46 AM
P: 610
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
#3
Mar9-12, 10:05 AM
P: 136
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
#4
Mar9-12, 11:02 AM
P: 610
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
#5
Mar13-12, 03:57 PM
P: 136
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