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Mechanics -straight line motion

  1. Mar 19, 2006 #1
    Im really puzzled on this question:
    A car intially accelerates at a(t) = 1/40 sin((pi x t)/1800) m/sec where t is time. What is the distance travelled in km when the car next comes to a halt. How long has the car been travelling? What is the maximum speed attained by the car?

    By integrating the acceleration for velocity and equating this to zero for when the train stops i found t=3600. When integrating again to give an equation for distance travelled by the car i got (45logt)/pi in the equation. Is this correct? If so how should the constant of integration in this equation be found given log0 is undefined? If log0 is taken as zero i get a distance travelled as 50.9m which seems too small. Am i going wrong somewhere?

    Also how would i go about finding max speed? Is this when acceleration is zero?
    Grateful for anyones help x
     
  2. jcsd
  3. Mar 19, 2006 #2

    Hootenanny

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    I think you've integrated wrong. I get t=900 for when the train stops.
     
  4. Mar 19, 2006 #3

    Hootenanny

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    You cannot take log0 as 0 because it is undefined! If you sketch the graph of y = logx it does not intersect the y-axis.
     
  5. Mar 19, 2006 #4
    Ok so i found the time to actually be 900 but i am still having problems finding the distance. When i integrate again to get distance i get
    -81000/(pi^2 *t^2)sin((pi*t)/1800), i think this is correct?
    However at t=900 the sin part comes out as 1, and so the distance is very small, ~0.01m. This is also a minus number, should the sign just be ignored because it is a distance? I think i am doing something wrong here, or just missing something obvious! Please help x
     
  6. Mar 19, 2006 #5

    Hootenanny

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    For the intergral of accleration I get;

    [tex]\int a(t) \;\;dt = -\frac{\pi}{72000}\cos\left( \frac{t\pi}{1800} \right)[/tex]

    Do you follow and agree?
     
  7. Mar 19, 2006 #6
    i get -45/(pi*t) as the bit before cos. i divided the 1/40 by (pi*t)/1800 is this wrong?
     
  8. Mar 19, 2006 #7

    Hootenanny

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    No you're almost right, I apologise, I thought I was differentiating again :frown:, time to get more coffee.

    So the correct verson is;
    [tex]\int a(t) \;\;dt = -\frac{45}{\pi}\cos\left( \frac{t\pi}{1800} \right)[/tex]

    You don't bring the t to the front when intergrating. I suggest you revise trig calculus.
     
    Last edited: Mar 19, 2006
  9. Mar 19, 2006 #8
    okay, but without t would there not also be a constant of integration in this equation? +45/pi ??? due to speed=0 at t=0, and cos0=1
     
  10. Mar 19, 2006 #9

    Hootenanny

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    There would indeed, you are quite right! My math's tutor is always on at me for my constants :(

    That is of course if you assume it is at rest.
     
    Last edited: Mar 19, 2006
  11. Mar 19, 2006 #10
    ok so with the constant i get that t=3600 when the car stops as cos of angle must =1 so the constant cancels out. I then get the distance travelled when the car stops as 51.6km. Do you agree?

    How do you tackle the final part of the question? Is maximum speed at a=0?
     
  12. Mar 19, 2006 #11

    Hootenanny

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    Yes I agree

    Yes. To check that it is a maximum [itex]a'(t) < 0[/itex] :smile:
     
  13. Mar 19, 2006 #12
    thanks very much for your help
     
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