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Given the following formulas

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that the position vector for a particle is given as a function of time by (t) = x(t) + y(t), with x(t) = at + b and y(t) = ct2 + d,where a = 1.70 m/s, b = 1.20 m, c = 0.122 m/s2, and d = 1.18 m.
    (a) Calculate the average velocity during the time interval from t = 2.05 s to t = 4.05 s.
    (b)Determine the velocity at t = 2.05 s.
    (c)Determine the speed at t = 2.05 s.

    2. Relevant equations
    for a.) I used Vavg=delta r/delta t


    3. The attempt at a solution
    for a.) for rf i got x(t)=8.05
    y(t)=3.181

    for ri i got x(t)=4.685
    y(t)=1.69

    and delta t=4.05-2.05=2

    Im not getting it. Someone please help
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 16, 2012 #2

    rl.bhat

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    Post the complete problem.
     
  4. Sep 16, 2012 #3
    listing the variables as a,b,c,d does us no good. We can assume what they are with the units but i don't know what b and d are.

    Basically, use the standard symbols.
     
  5. Sep 16, 2012 #4
    sorry about that
     
  6. Sep 16, 2012 #5

    Simon Bridge

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    Do I take it that: $$\vec{r}(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}$$... where: ##x(t)=v_xt+x_0## and ##y(t)=a_yt^2+y_0##

    These are position-time functions - I changed the variable names to reflect their roles.
    From these you can find ##v_x(t)##, and ##v_y(t)##

    (a) average velocity is change in position over change in time - you have two position vectors - what is the distance between their end-points?

    (b) ##v_x(t)=\dot{x}(t)##

    (c) how is the speed and the velocity related?
     
  7. Sep 16, 2012 #6

    SammyS

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    Do you know Calculus?
     
  8. Sep 17, 2012 #7

    Simon Bridge

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    Gudduy cobber;
    Does not need calculus if kinematics is available
    ... anyway, piercegirl got stuck on average velocity. The calc is in the next bit.

    I know - calculus is fun and we want to spread the Word... ;)
     
  9. Sep 17, 2012 #8

    I like Serena

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    Hi piercegirl!

    You're on the right track.
    I guess you need to be aware of the fact that an average "velocity" is a vector.

    You have position vectors at both locations.
    Can you subtract those position vectors, yielding a change in position vector?
    And divide that by the change in time?

    In other words, can you calculate:
    $$\mathbf{\vec v}_{avg} = {\mathbf{\vec{Δr}} \over Δt}$$
     
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