Recent content by Brett R.

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    Why Must the Normal Force Equal Zero at the Top of a Loop?

    Homework Statement In the normal loop the loop problem involving rotational energy where the marble goes down the hill and goes through a loop the loop, it asks for the minimum height of the hill to keep the marble on the track. Homework Equations But why does the normal force have to equal 0...
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    Physics and Accident Reconstruction: Solving the Drag Equation

    Is it 1/1+(kv)2? That is my goal, however I don't think this equation will tell me that goal anymore.
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    Physics and Accident Reconstruction: Solving the Drag Equation

    However, I want to know how drag changes with velocity. How will this help me with that?
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    Physics and Accident Reconstruction: Solving the Drag Equation

    So... I think I got it. So the integral is arctan(kv) if you set β/μg = k2 which makes things a lot easier. Then you substitute back in for k and you get arctan(βv/μg)=0 right?
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    Physics and Accident Reconstruction: Solving the Drag Equation

    However, I need the form 1/1+v2. I have 1/(1+ (β/μg)v2). How do I get that other part out.
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    Physics and Accident Reconstruction: Solving the Drag Equation

    Sorry for all the questions, but my math skills have not gotten me to the point where I can take all the information I learn and apply it to physics equations. I really appreciate your help.
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    Physics and Accident Reconstruction: Solving the Drag Equation

    So it is arc tan. i manipulate to get ∫dt= (1/μg)[1/(1+(β/μg)v2)]dv
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    Physics and Accident Reconstruction: Solving the Drag Equation

    OHH I think i got it. Because nothing here but velocity varies with time. So if I just divide, I can integrate dt on the left and 1/the rest on the right
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    Physics and Accident Reconstruction: Solving the Drag Equation

    that would leave it in the form. (-μg)dt + (-βv2) dt=dv This is where I was stuck. I do not know how to go farther without further complicating the terms.
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    Physics and Accident Reconstruction: Solving the Drag Equation

    I know but if you divide by both sides, you are left with an m under the second term. Then I substituted out. As for dt, if I multiply across I can not get the v^2 to the right since it would be trapped by the dt.
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    Physics and Accident Reconstruction: Solving the Drag Equation

    I divided both sides by m to get the m on the right side to go away. My problem is I was having trouble getting the v^2 term to the right
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    Physics and Accident Reconstruction: Solving the Drag Equation

    how do I integrate -μg-βv2 = dv/dt
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    Physics and Accident Reconstruction: Solving the Drag Equation

    Well I am using a Skid mark equation √μgdcosθ. So a car is traveling along the x and skids to a stop. I want to compare the various effects of velocity and frontal area on drag. I want to know if drag will have a major effect on the calculated velocity based on a skid mark length. So there is...
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    Physics and Accident Reconstruction: Solving the Drag Equation

    HI all, I am new to the forum and have a real question for you physics buffs. It has to do with drag. For a mathematics paper, I am exploring the use of physics in accident reconstruction. I want to use calculus to create differential equations for the drag equation .5p(Cd)(A)v^2... I want a...
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