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!

Kinematics of Particles

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle moves in a plane. The particle (x,y) position on the plane is given by:

    [tex]x = -34t^4 − 28t^3 + 7[/tex]
    [tex]y = 25t^2 + 13t^3 + 5[/tex]

    Therefore the particle's displacement [from the (x,y) origin) at time t = 68 s is

    [tex](736000000m) i + (4200000m)j[/tex]

    (a) What is the particle's velocity at 68 s ?

    (b) What is the particle's acceleration at 68 s ?

    3. The attempt at a solution

    (a) for this part the correct answer has to be

    [tex]v_p = (43200000m) i + (184000m)j[/tex]

    But I can't see how they have arrived at this answer! The velocity is [tex]\frac{\Delta x}{\Delta t}[/tex]

    [tex]\Delta t = t_f-t_i=68-0 =68[/tex]

    [tex]\Delta x = x_f-x_i = 736000000 - 7 = 375999993[/tex]

    So, [tex]\frac{375999993}{68}=10823529.31[/tex].

    This is not the right answer for the i component. What is the problem?
     
  2. jcsd
  3. Mar 22, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    just use v(t)= (dx/dt)i+(dy/dt)j
     
  4. Mar 22, 2010 #3
    That's exactly what I did! And I don't know why I get the wrong answer...
     
  5. Mar 22, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    Can you show exactly how you did it? You may have made a mistake in the algebra somewhere.
     
  6. Mar 22, 2010 #5
    Here is my working for the i component:

    The particle's displacement is [tex](736000000m) i + (4200000m)j[/tex]

    [tex]\Delta t = t_f-t_i=68-0 =68[/tex]

    Since for t=0 the first equation, [tex]x=-34t^4-28t^3+7[/tex], will be 7.

    [tex]\Delta x = x_f-x_i = 736000000 - 7 = 375999993[/tex]

    So, [tex]v=\frac{\Delta x}{\Delta t}=\frac{375999993}{68}=10823529.31[/tex].
     
  7. Mar 22, 2010 #6

    rock.freak667

    User Avatar
    Homework Helper

    In your answers are you working with a specified degree of accuracy?

    Also if x=-34t4-28t3+7, how is your i component positive?

    Alos what is dx/dt and dy/dt equal to in terms of t?
     
  8. Mar 22, 2010 #7
    I think 2% is the accuracy tolerance.

    oops, I'm sorry, the answer has to be -43200000 i. :redface:

    [tex]\frac{375999993}{68}=10823529.31[/tex]

    If this not true then I have no idea...
     
  9. Mar 22, 2010 #8

    rock.freak667

    User Avatar
    Homework Helper

    If x=t5, then dx/dt is 5t4


    So if x=-34t4-28t3+7 and y =25t2+13t3+5

    what is dx/dt and dy/dt equal to?
     
  10. Mar 23, 2010 #9
    Thanks a lot it worked! :)

    Finally, the question asks "what is the particle's acceleration at 68 s?". If I divide the particle's velocity at 68 s by time=68 seconds, it doesn't give me the correct answer. So what formula do I need to use?
     
  11. Mar 23, 2010 #10

    rock.freak667

    User Avatar
    Homework Helper

    You use the same method.


    a(t)=(d2x/dt2)i+(d2y/dt2}j
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinematics of Particles
  1. Kinematics of particles (Replies: 21)

  2. Kinematics of a particle (Replies: 10)

Loading...