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!

Kinematics and acceleration problem

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle moves along a straight line such that its acceleration is a = (4t2-2) [itex]\frac{m}{s^2}[/itex], where t is in seconds. When t = 0, the particle is located 1 m to the left of the origin, and when t = 2 s, it is 20 m to the left of the origin. Determine the position of the particle when t = 4.3s.

    2. Relevant equations & The attempt at a solution.

    I am just stuck on this. I know I should already know how to do this but right now my mind is blank!
    I know that I can get the velocity equation by integrating, but when I try to use t=2 s, and then using the equation x=xo+vot+[itex]\frac{1}{2}[/itex]at2, I am not even close to the x=20.
     
  2. jcsd
  3. Jan 23, 2013 #2

    tms

    User Avatar

    The kinematics equation you mention is good only for constant acceleration, but the acceleration in the problem varies with time, so you have to go back to basics. Start with the definition of acceleration: [itex]a = dv/dt[/itex].
     
  4. Jan 23, 2013 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That equation is only valid for constant acceleration.

    What do you get when you integrate the acceleration equation?
     
  5. Jan 23, 2013 #4
    After I integrate the acceleration, I get:

    v(t)= ([itex]\frac{4}{3}[/itex]t3 - 2t) [itex]\frac{m}{s}[/itex]
     
  6. Jan 23, 2013 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Don't forget the constant of integration.
    Then get the equation for distance as a function of time.
     
  7. Jan 23, 2013 #6

    tms

    User Avatar

    Don't forget the constant of integration.

    Then the next step is to find the displacement.
     
  8. Jan 25, 2013 #7
    s(t) = [itex]\frac{t^4}{3}[/itex] - t2 + c1t + c2

    c1 is the constant after integrating for v(t) & c2 is the constant for s(t).

    Do I have to set the constants equal to the given positions and t=0 and t=2?
     
  9. Jan 25, 2013 #8

    tms

    User Avatar

    No. You plug in the value for [itex]t[/itex], and then solve for the constants.
     
  10. Jan 26, 2013 #9
    Ok. I plugged in the t=0, s=1 and got c2 = 1 (I think this would be -1 since the location is to the left of the origin?)

    So I then plugged t=2, s= -20 and my c1 = -9.84

    Finally I plugged my c1 and c2 into my s(t) equation:

    s(t) = [itex]\frac{1}{3}[/itex]t4 - t2 - 9.84t -1

    And after using t=4.3, I got s = 52.16 m

    Unfortunately, this is not correct. The correct answer was 50.8 m. I'm not sure where I messed up, but I'm thinking it was my rounding.
     
    Last edited: Jan 26, 2013
  11. Jan 26, 2013 #10

    Found my mistake!! I was subtracting my numbers from +20 and not using -20 when calculating my c1!!

    I got:
    s = 50.75 = 50.8 m!!

    Dang!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinematics and acceleration problem
Loading...