1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Basic Velocity -> Speed -> Position of particle

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle's trajectory is described by x=((1/2)*(t^3)-2t^2)m and y=((1/2)*(t^2)-2t)m, where t is in s.

    a) What are the particle's position and speed at t=0s and t=4s?

    b) What is the particle's direction of motion, measured as an angle from the x-axis, at t=0s and t=4s?

    2. Relevant equations
    Listed above

    3. The attempt at a solution
    I feel like I am over-thinking the problem. As I understand it the equations provided are velocities in the directions of the x and y axies. So, the integral of the equation for 'x' would give me the speed. Integrating it a second time would give me position. I can work through the integrals but something doesn't feel right about the relationship between the work I'm doing and the material in the chapter.

    integral of x = 1/8 t^4 - 2/3 t^3 and second integral of x = t^5/40 - t^4/6

    integral of y = 1/6 t^3 - t^2 and second integral of y = t^4/24 - t^3/3

    Plugging in the values of 0 and 4:

    The position at 0 seconds = (0,0)
    The speed at 0 seconds = 0 m/s

    The position at 4 seconds = (-17 1/15, -10 2/3)
    The speed at 4 seconds = -10 2/3 m/s

    The second part I can simply use the triangles created to find the angle in relation to x.

    Am I on the right track?
  2. jcsd
  3. Sep 11, 2011 #2


    User Avatar
    Homework Helper
    Gold Member

    Workin' in the wrong direction (differentiate, not integrate). dx/dt = v, dv/dt = a
  4. Sep 11, 2011 #3
    Bah, of course. Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook