Motion in 2d/Centripetal Acceleration

  1. ...sorry for doing this, but having trouble with some problems

    1) A particle starts from rest at t=0 at the origin and moves in the xy plane with a constant acceleration of a= (2i + 4j) m/s2. After a time t has elaped, determine (a) the x and y components of velocity. (b) the coordinates of the particle, and (c) the speed of the particle.

    I get a) Vx= (2t)m/s and Vy=(4t)m/s, but not the rest...


    2) A ball swings in a vertical circle at the end of a rope 1.5 M long. When it is 36.9* past the lowest point on its way up, the ball's total acceleration is (-22.5i + 20.2j) m/s2. For that instant, a) determine the magnitude of its centripetal acceleration, and b) determine the magnitude and direction of its velocity.

    I know that v^2/R= Ac, but i don't know how to get the velocity:(

    Any help will be appreciated
  2. jcsd
  3. Pyrrhus

    Pyrrhus 2,273
    Homework Helper

    Use uniform acceleration kinematic equations, or use calculus. Integrate Acceleration, then integrate Velocity.

    [tex] \vec{a} = \frac{d \vec{v}}{dt} [/tex]

    [tex] \vec{v} = \frac{d \vec{r}}{dt} [/tex]


    [tex] A_{c} = \frac{v^2}{r} [/tex]

    Refers to Magnitude.
  4. but how do i find out the velocity?
  5. Well I got it. For 1 b) x=t^2m y=2t^2, for c) a^2 + b^2= c^2, c= velocity.

    For 2. The angles of the directionals were 36.9 and 53.1, then cos36.9=20.2/x, solve for x, cos53.1= 22.5/x2, solve for x2, add both xs, and you get 62.8 m/s for the magnitude of its centripetal acceleration. For c, v^2/r= aC, we have the aC and r, so v^2/1.5=62.8, simple algebra, and v=9.7m/s.
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