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What is the acceleration of the particle

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data
    The motion of a particle is given as follows :
    ##x=u(t-2)+a(t-2)^2##
    Which of the following option(s) is/are correct ?
    A.The initial velocity of particle is u
    B.The acceleration of the particle is a
    C.The acceleration of the particle is 2a
    D.At ##t=2s## particle is at origin.

    2. Relevant equations
    ##x=ut+\frac{1}{2}at^2##

    3. The attempt at a solution
    Simplifying the equation, we get,
    ##x=ut-2u+at^2-4at+4a##
    Could somebody guide me as to what will be the next step ?
     
  2. jcsd
  3. Dec 4, 2014 #2
    Have you had (or are studying) calculus?

    Chet
     
  4. Dec 5, 2014 #3
    No. I have had no exposure to calculus.
     
  5. Dec 5, 2014 #4
    Well, how are velocity and acceleration defined and related to each other?
     
  6. Dec 5, 2014 #5
    The mentioned scenarios can be reduced to relative motion. Right ?
     
  7. Dec 5, 2014 #6
    Nope.
     
  8. Dec 5, 2014 #7
    Well, without calculus, it's not as easy , but try this: substitute t'=t-2 in your original equation and see shat you get.

    Chet
     
  9. Dec 5, 2014 #8
    But there is no t'. I just simplified the problem equation.
     
  10. Dec 5, 2014 #9
    Let's try something different. Let's take your simplified equation and rewrite it as:
    [tex]x=(4a-2u)+(u-4a)t+at^2[/tex]
    Compare this with:
    [tex]x=x_0+v_0t+\frac{1}{2}At^2[/tex]
    where
    x0=particle location at time zero
    v0=particle velocity at time zero
    A = particle acceleration

    Chet
     
  11. Dec 6, 2014 #10
    dx/dt=v
    dv/dt=a
    I think you may be familiar with derivatives even if you haven't had much exposure to calculus.
     
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