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Thread: Velocity of a particle

  1. #1
    Apr 2008

    Velocity of a particle

    Ok i have had enough of failing to finish this one so here is the question and I will provide any work done below it. Long story short I am still trying to figure out the velocity of the partical at it's point of release from the circle. Only real big hint i can provide (according to my instructor) is the value will be something around 500.000XXXXXXXXXX

    A particle is being accelerated by a hypothetical cyclotron in an eletric/magnetic field such that it follows the path indicated by the vector curve $\displaystyle r(t)=(0.50000cost^2)i+(0.50000sint^2)j$ where t is in seconds and r(t) is in miles. When the field is turned off, the particle is to fly off tangentially to hit a target that stands at the coordinates (-1,0) miles (the vector -i) based on the coordinate system of r(t). The field is to be turned off the first time the particle would fly off to hit the target after the speed reaches 500 miles/second. The vector sum that would then be equivalent to the vector -i is r(t)+pv(t) for some scalar p. Correct to 15 significant digits, identify the actual speed of the particle just befor it hits the target (it has a constant speed of flight).

    Now for what I have figured

    $\displaystyle r(t)=<.5cos(t^2),.5sin(t^2)$=position vector
    $\displaystyle r'(t)=v=<-tsin(t^2),(t)cos(t^2)>$=Velocity vector
    $\displaystyle |r'(t)|=|v|=sqrt(((t)cos(t^2))^2+((t)sin(t^2))^2)$=speed

    by setting the |v| = to 500 i found the value of t to be 500 so t=v therefor speed and time are the same value

    At that position in which the particle reaches the speed of 500miles/second it is at $\displaystyle ((cos250000)/2,(sin250000/2)$ thus putting it in the third quadrent. The point in which it needs to release is located in the second and therefor the partical needs to continue along the path of the circle (and thus still speeds up) till it reaches the point in which it would fly off to hit the point (0,-1)

    without making a long old detailed reason as to how i came to it I figured the point in which it would release from the circle to be at

    $\displaystyle (-1/4,sqrt(3)/4)$= the point p

    So aside from that I am stuck on the final determination of how to figure the final velocity value when it reaches that point. Any help would be appreciated.
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  2. #2
    MHF Contributor
    skeeter's Avatar
    Jun 2008
    North Texas
    here's a stab at it ...

    .5\cos(t^2) = -\frac{1}{4}

    \cos(t^2) = -\frac{1}{2}

    $\displaystyle t^2 = \frac{2\pi}{3} + 2k\pi > 500^2$

    $\displaystyle 2k\pi > 500^2 - \frac{2\pi}{3}$

    $\displaystyle k > \frac{500^2}{2\pi} - \frac{1}{3}$

    k > 39788.40244

    since $\displaystyle k$ is an integer ...

    $\displaystyle k = 39789$

    $\displaystyle t = \sqrt{\frac{2\pi}{3} + 2 \cdot 39789 \cdot \pi}$

    my calculator will not give 15 significant digit accuracy
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