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!

Deriving acceleration from a position function

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle moves along the y axis according to y(t)=(3.5 m/s2)t2 - (9.0 m/s)t + 5.0 m.
    (a) What is the particle's velocity as a function of time?
    (b) What is the particle's acceleration at t=2.0 s?

    3. The attempt at a solution

    I was wondering about b- I derived:
    dy/dt = (7 m/s2)t - 9.0 m/s for the velocity function
    and then
    dv/dt = 7 m/s2 for the acceleration function.

    So the acceleration is constant, right? I'm finding seemingly irrelevant questions make me paranoid.
  2. jcsd
  3. Aug 31, 2010 #2
    Yes, you just showed that the acceleration is constant. In fact you can check with a graphing calculator by applying the knowledge you learned in class.
    In fact when you graph the y(t) equation you get a upwards parabola. Then the v(t) graph will be a positive line. Thus the a(t) graph is constant.
    My calculator says you are right. :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook