Physics Forums (http://www.physicsforums.com/index.php)
-   Introductory Physics Homework (http://www.physicsforums.com/forumdisplay.php?f=153)
-   -   rectilinear motion (displacement, position) calculus (http://www.physicsforums.com/showthread.php?t=585102)

 xzibition8612 Mar8-12 05:03 PM

rectilinear motion (displacement, position) calculus

1. The problem statement, all variables and given/known data
A particle has a linearly varying rectilinear acceleration of a=x''i=(12t)i m/s^2. Two observations of the particle's motion are made: Its velocity at t = 1s is x'i=2i m/s, and its position at t= 2s is given bt xi=3i m.
(a) Find the displacement of the particle at t=5s relative to where it was at t = 0s.
(b) Determine the distance traveled by the particle over the same time interval.

2. Relevant equations

3. The attempt at a solution
Given:
x''=12t
x'(1)=2
x(2)=3

Integrate x''=12t and apply initial conditions and get the equations of motion:
x''=12t
x'=6t^2-4
x=2t^3-4t-5

(a) plug in: x(5)-x(0) and get 230m. Correct answer
(b) Isn't it the same thing? x(5)-x(0)? But the book says its 234 m. No idea how this came about.

 OldEngr63 Mar8-12 07:09 PM

Re: rectilinear motion (displacement, position) calculus

To get the distance traveled, calculate

dist = ∫ |x'| dt

=∫05 | 6t2 - 4 | dt

=∫0√(2/3) (-6t2 + 4) dt

+∫√(2/3)5 (6t2 - 4) dt

= 2.1773 + 232.1773

= 234.354

The difference is that this motion reverses direction.

 xzibition8612 Mar9-12 10:10 AM

Re: rectilinear motion (displacement, position) calculus

so distance is the integral of velocity? Ok, now I don't get where the sqrt(2/3) came from. What's that? And why do you take the absolute value of velocity?

Thanks.

 OldEngr63 Mar9-12 11:17 AM

Re: rectilinear motion (displacement, position) calculus

The distance is the integral of the absolute value of the velocity. The distance is the total amount of motion (tire wear, if you want to look at it that way). This system starts off moving to the left until t = sqrt(2/3) at which time the velocity goes to zero and the motion reverses. Then the system moves to the right from t = sqrt(2/3) until t = 5.

To start the problem, you have to recognize that the motion will reverse, and you have to find the time at which that reversal occurs. Then you work the problem in two parts, before and after the reversal.

 xzibition8612 Mar13-12 05:12 PM

Re: rectilinear motion (displacement, position) calculus

How did you get t=sqrt(2/3)? Is it from the equation 2t^3-4t-5=0? If so can you show the steps because I can't figure it out for the life of me this cubic equation.

 OldEngr63 Mar13-12 08:22 PM

Re: rectilinear motion (displacement, position) calculus

It comes from finding the zeroes of the velocity function. This is necessary to understand how to handle the sign changes required for the absolute value function.

 All times are GMT -5. The time now is 06:29 AM.