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Please help w/ a position-vector question

  1. Jan 25, 2008 #1
    Suppose that the position-vector function 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 = 1.40 m/s, b = 0.75 m, c = 0.120 m/s2, and d = 1.10 m.

    (a) Calculate the average velocity during the time interval from t = 1.75 s to t = 3.85 s.
    ( ___ î + ___ ĵ ) m/s

    (b) Determine the velocity at t = 1.75 s.
    ( ___ î + ___ ĵ ) m/s

    Determine the speed at t = 1.75 s.
    ___ m/s

    The answers I got for the first four slots are: 3.2, 1.46, 1.4, and 0.42. I don't know if those values are correct. And can someone please teach me how to do part 5? Thank you.

    x(t) = 1.4(t) + 0.75
    y(t) = 0.12(t)^2 +1.1
     
  2. jcsd
  3. Jan 25, 2008 #2
    Substitute the value of t as 1.75 & differentiate the equation wid respect to time and then ya get the velocity....
     
  4. Jan 25, 2008 #3
    So for part a...
    1.4(1.75) + 0.75 = 3.2
    &
    0.12(1.75)^2 +1.1 = 1.46

    For part b, after differentiating I get...
    1.4
    &
    0.24(1.75) = 0.42

    Is that correct?

    The part I'm most confused about is the finding the speed for the last part. I know it has to do with magnitude.
     
  5. Jan 25, 2008 #4
    ur correct...now take modulus of velocity to get the speed....or in other words magnitude as ya say...
     
  6. Jan 25, 2008 #5
    Opps, for part a.) would I have to divide those values by 2.1? Since it says during the time interval betw. t = 1.75 s to t = 3.85 s.

    And can you describe to me how to go about finding the magnitude please?
     
  7. Jan 25, 2008 #6
    of course u would have to divide it by 2.1...

    now for magnitude..

    magnitude i.e.speed= {[velocity along x axis]^2+ [velocity along y-axis]^2}^1/2
     
  8. Jan 25, 2008 #7
    Thank you very much
     
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