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Homework Help: Dynamics: Curvilinear motion

  1. Feb 9, 2008 #1
    Hi guys,Im Chris, im new here. Im in Dynamics doing curvilinear motion.
    1. The problem statement, all variables and given/known data

    A truck travels at a speed of 4m/s along a circular road with radius if 50m. For a short distance, from s=0, its speed is then increased by a(tan) = (0.05s) m/s, where s is in meters. Determine the speed and acceleration magnitude at s=10. (i can find the a-magnitude easily after i find v, i just cant find v)

    2. Relevant equations
    acceleration and velocity equations
    a(tan)ds = v dv

    3. The attempt at a solution
    Ive tried to attempt this using integration with respect to s, but this does not yield the correct answer, so its the incorrect procedure. The answer is 4.58 m/s.
    I have never seen or dealt with integration with respect to displacement, therefore i dont know how to approach this problem. The book says nothing on figuring this out. Any help is appreciated, Thanks!
    Last edited: Feb 9, 2008
  2. jcsd
  3. Feb 9, 2008 #2

    Tom Mattson

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    Science Advisor
    Gold Member

    Hello, and welcome to PF!

    Integration is indeed the way to go, so you must have made an error there. Let's see what you did, and then we can help you.
  4. Feb 9, 2008 #3
    Thank you! well i integrated and got v = 0.025s^2, s = 10. This yeilds 2.5 m/s. Adding this to the initial 4 m/s yields 6.5 m/s. This is incorrect. the answer is 4.58 m/s. So, where is my error? I am quite dumbfounded...
  5. Feb 9, 2008 #4
    Im starting to think this should be in advanced physics as it requires perhaps some difficult manipulation? Im in calc 3 and have no idea what to do...
    To the Moderator, please move if you feel fit
    Last edited: Feb 9, 2008
  6. Feb 10, 2008 #5
    You can not add the initial velocity. From the integration you have

    [tex]\int_4^{v_f} v\,d\,v=\int_0^{s_f}\frac{s}{20}d\,s\Rightarrow \frac{1}{2}\,v^2\Big|_4^{v_f}=\frac{1}{40}\,s^2_f \Rightarrow u_f^2-4^2=\frac{1}{20}\,s_f^2[/tex]

    Plug your values in the above equation to find the answer.
  7. Feb 10, 2008 #6
    wow,i did not even think about integrating the v from 4-vf. Thank you so much. But just to be clear, you say that i cant add the initial velocity. The final answer of that integration is .58m/s so that would conclude that I do have to add the initial 4 m/s to get 4.58m/s. Isnt this correct?
    Last edited: Feb 10, 2008
  8. Feb 10, 2008 #7
    Which integration? :confused:
    If the initial velocity is [itex]v_i[/itex] then the final velocity is [itex]v_[/itex] then

    [tex]v_f^2=v_i^2+\frac{1}{20}\,s^2\Rightarrow v_f=\sqrt{v_i^2+\frac{1}{20}\,s^2}[/tex]

    The final velocity is not proportional to the initial velocity. That's why you can not add it.

    Is it clear? :smile:
  9. Feb 10, 2008 #8
    Hmm, now im even more confused. Plugging s and vi into that equation gives me 6.4 m/s = vf. I dont understand, if the initial vi is 4m/s, and then accelerates for 10m to a new vf, wouldnt you be adding vi + delta v to get final velocity?

    [tex]u_f^2-4^2=\frac{1}{20}\,s_f^2[/tex] solving for uf yields .58
    adding 4+.58 yeilds the correct answer
    Last edited: Feb 10, 2008
  10. Feb 10, 2008 #9
    How did you find [itex]v_f=6.4\,m/sec[/itex]? For [itex]s=10\,m,\,v_i=4\,m/sec[/itex] you get

    [tex]v_f=\sqrt{16+\frac{100}{20}}\Rightarrow v_f=\sqrt{21}\Rightarrow v_f=4.58\,m/sec[/tex]

    The definition of the acceleration is [itex]a=\frac{d\,v}{d\,t}\Rightarrow d\,v=a\,d\,t\Rightarrow v_f-v_i=\int_0^{t_f}a\,d\,t[/itex]. If you call [itex]\Delta v=\int_0^{t_f}a\,d\,t[/itex] then of course you can write [itex]v_f=v_i+\Delta v[/itex].
  11. Feb 10, 2008 #10
    oh yes, your right, i was squaring the 100/20 by accident. I still dont fully understand the entire process but ill try and figure it out. Thanks again Rainbow Child.
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