1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constant acceleration motion problem

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle moving along a straight line with constant acceleration starts its motion at t = 0. The particle is observed to change direction at t=4s, and when t=10s it reaches the position of -2 m with a velocity of -2.4 m/s measured from a chosen reference frame. Find the position of the particle when it changes direction of motion.

    Answer: 5.2 m


    2. Relevant equations

    x = x0 + v0*t + (1/2)at^2
    v = v0 + at

    and possibly: v^2 = v0^2 + 2a(x-x0)

    3. The attempt at a solution

    I'm not really sure how to go about doing this problem. I'm getting a little confused on the defining variables part even -- this is what I THINK they should be defined as but I'm not sure:
    x = -2 m at t=10
    x0 = ?
    v = 0 m/s at t=4s, -2.4 m/s at t=10
    v0 = ?
    a = ?

    It seems like I'm missing 3 things. I'm thinking either v0 or x0 is 0, but I'm not sure which of those is 0.

    I think I'll only be using the first 2 equations since time is involved, and that's really as far as I got on this problem. :( Any help is appreciated, I have a quiz that includes this problem due tomorrow at 1:25 PM eastern time.
     
  2. jcsd
  3. Feb 3, 2010 #2
    Someone can close this -- I figured it out on my own. Used systems to solve this one per a friend's tip. Solution is below for anyone interested:

    0 = v0 + 4a
    -2.4 = v0 + 10a
    ==============
    2.4 = -6a
    a = 2.4/-6 = -0.4

    0 = v0 + 4 * -0.4
    v0 = 1.6

    -2 = x0 + 1.6 * 10 + .5 * -0.4 * 100
    x0 = 2

    x = 2 + 1.6 * 4 + .5 * -0.4 * 16
    x = 5.2 m
     
  4. Feb 8, 2011 #3
    How did you get the -6a?
     
  5. Feb 8, 2011 #4
    I just subtracted the two equations. That's why the 2.4 is positive on the lefthand side and the v0 disappears.

    Also, this thread is OLD and has been solved already. Someone can close it as I said in the previous post
     
  6. Feb 8, 2011 #5
    I got a similar question on a quiz due tomorrow, Chang ftw
     
  7. Feb 8, 2011 #6
    Haha nice, you're a VT student too? I had Chang a couple semesters ago...I thought he stopped teaching?
     
  8. Feb 8, 2011 #7
    Ya, second semester at tech. Chang still teaches, I was pretty lucky to get him haha
     
  9. Feb 8, 2011 #8
    Yeah, honestly, all the Physics professors EXCEPT for Chang are horrendously bad at VT. Enjoy 2306 next semester...it's a fun time.
     
  10. Feb 8, 2011 #9
    Ya I heard, thanks for the help. I have to figure out the rest of the stuff haha
     
  11. Feb 8, 2011 #10
    No problem
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook