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!

At what times are the velocity and acceleration perpendicular?

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data

    The coordinates of the particle is given by:

    x(t)=(1.2 m/s)t
    y(t)=15m-(0.5m/s2)t2

    a) At what times is the particle's velocity perpendicular to its acceleration
    b) At what times is the particle's speed instantaneously not changing (I get that the acceleration is zero).
    c) At what times is the particle's position perpendicular to its velocity.
    d) What is the particle's minimum distance from the origin.


    2. Relevant equations

    r=xi+yj (i and j are the corresponding unit vectors)
    vx=dx/dt
    vy=dy/dt
    a=dv/dt.

    3. The attempt at a solution

    I tried to differentiate the particle position r with respect to time getting velocity, and then taking the derivative of the velocity with respect to time getting acceleration.
    When I had the acceleration as function of time, I got: a=(-2*0.5 m/s2)j. But how can I then solve for t, when it is not in the equation? What have I done wrong? Please help me with all the questions.
     
  2. jcsd
  3. Aug 20, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    j is the y-direction? Okay.
    So now you can insert the requirement "acceleration is orthogonal to the velocity". If the acceleration points in y-direction, which direction do you need for the velocity?
    This will lead to an equation with time inside.
     
  4. Aug 20, 2013 #3

    verty

    User Avatar
    Homework Helper

    For part a, use the dot product. For part b, think about centripetal acceleration, the speed can be constant even though there is acceleration. The condition you want is, the velocity and acceleration are orthogonal (perpendicular with the addition that the zero vector is orthogonal to every vector). The rest I'll leave to you.

    Got beat by mfb but I think both replies are useful.
     
  5. Aug 20, 2013 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are you unable to simplify -2*.05? Just asking..

    So the object has constant, nonzero, acceleration. There is no time t where the acceleration is zero, so that's why you can't make it work.

    If you want help on the other parts, you need so show your work and explain where you are stuck.
     
  6. Aug 20, 2013 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Never, never include units like that in your equations; just write
    x(t) = 1.2*t and y(t) = 15 -.5*t2 (in m/sec). For example, if you tried to submit your equations---exactly as written---to a computer algebra package, the computer would choke, or else would severely misinterpret your formulas.
     
  7. Aug 20, 2013 #6

    verty

    User Avatar
    Homework Helper

    Some books teach that as a way to catch errors, it seems. Mistakes in calculations are pernicious, so it may be a useful investment of time in tests where one is usually under stress anyway.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted