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!

Dynamics with complex equations

  1. Sep 16, 2010 #1
    Problem:
    A particle starts moving from the position
    r= -2i+j

    with its velocity given as a function of time, t, as

    v=(5t^2)i+2j
    .
    What is the distance of the particle from the origin at time t = 10 s?

    The i's a j's both have hats if that makes any difference.

    My attempt:
    distance = velocity x time
    so multiply the v equation by t and then subsitute 10s into it.
    This will give the distance travelled in both the x and y direction add these values to the values given in the first equation.
    So (5t^2)*t + t*2j

    then sub in 10 and add the i and j values to their respective values in the first equation to find the final displacement.

    I'm unsure if this is correct or not.
     
  2. jcsd
  3. Sep 16, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No, "displacement", not "distance" equals velocity x time, and even that is only true in the case of constant velocity.

    Instead, you will want to integrate the velocity to get the position as a function of time, and then determine the distance from the origin at a given time by using the usual formula for the distance between 2 points.

    [/QUOTE]
     
  4. Sep 16, 2010 #3
    is this not exactly what you're telling me to do?
    (5t^2)*t + t*2j
     
  5. Sep 16, 2010 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No, do you know how to integrate?
     
  6. Sep 16, 2010 #5
    Sorry didn't read what you first said properly.
    So once the integral is found, and the equation becomes something which represents the displacement of the object, can't I just sub 10 into this equation and then add -2i and 1j to the resulting i and j values?
     
  7. Sep 17, 2010 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member


    Sort of, your integration (like any integration) will have a constant on the end. In this case, the constant will be a vector and equal to the initial position of the particle:

    [tex]\int_0^t \textbf{v}(t')dt' = \int_0^t \frac{d\textbf{r}(t')}{dt'}dt' = \textbf{r}(t)-\textbf{r}(0)[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dynamics with complex equations
  1. Dynamical equations? (Replies: 4)

Loading...