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Homework Help: Graphing expressions without given values?

  1. Apr 28, 2008 #1
    This isn't really a homework problem. I'm studdying for the AP Physics C: mechanics test by working on practice free response problems that my teacher gave me. My question pertains to a problem in the 1993 set of questions.

    Let's say, for example, a car moves to the right with a constant force F and there's an opposing force of -kv where v is velocity and k is a constant.

    If it's given that v(0) = 0, and I have found an expression for velocity as a function of time in terms of F, k, and m, but I haven't been given values, and I'm suppose to graph v(t)... do I just make up numbers for F,k, and m? I don't think I've ever graphed anything without knowing values. Or am I just suppose to estimate what the curve will look like?

    Even if my expression is wrong, they'll probably still give me points for correctly graphing my wrong answer. That's how AP tends to work.

    Also, one other question: my Physics teacher has pretty much left out the calculus side of everything we've learned this year. Will this hurt me on the test? If so, what calculus should I study?
  2. jcsd
  3. Apr 28, 2008 #2
    You can get a rough idea of how the function will look like. The way I see this is by knowing some calculus at least so you can get the idea of how derivatives look like.

    For example, if you have a displacement function that is a simple quadratic, then you can make a rough graph of the velocity function since it's merely a linear function.

    Thus, if you have a function X(t) which measures displacement over time t, and you're given a graph instead of actual numbers, then you can apply some knowledge of what you know about derivatives. For example, if you have a max or min in the graph you're dealing with, then at that time 't' of V(t) where X(t) has a min or max, you should have a velocity of 0 since the tangent line is horizonal to the graph at the min/max of X(t).
  4. Apr 28, 2008 #3


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    Well, as far as this point is concerned I'd say yes. Because it's a calculus based physics AP problem, you can count on at least two out of six free response questions dealing in one way or another mainly with calculus. Whether this involves solving differential equations, evaluating integrals, or whatever varies from year to year but these will show up. Also, basic calculus knowledge can sometimes help simplify problems. In terms of what calculus to study, I'd say just basic differential and integral calculus. They don't put anything too difficult on there, maybe in integrals up to u substitutions. If you have a TI-89 you can probably cruise through without any knowledge cause the calculator will do it all for you! :rofl:
  5. Apr 28, 2008 #4
    Hm. I've never used a TI-89. I'll look into it, but I probably won't need one. It takes me longer to punch stuff in.

    I feel sorry for the kids in my class that aren't in calculus but are taking the test. I guess it's not really our teacher's job to teach us calculus.

    So when a problem says "derive an expression", am I pretty much expected to use calculus? Like, if I have an expression for acceleration and I'm asked to derive an expression for velocity? Or if I'm dealing with force and potential energy?
    Last edited: Apr 28, 2008
  6. Apr 28, 2008 #5
    Well, you'd have a much easier time deriving certain expressions with the use of calculus. For example, if you're given an acceleration of lets say, a(t) = 2t, then using calculus, you can get the indefinite integral and that's v(t) = t^2 + C. If you're given initial conditions, then you can solve for C. For example, if they give you a(t) = 2x, and they want you to solve for a velocity for t=2 where v(0) = 2, then one finds that C = 2 and v(2) = 2^2 + 2 = 4 + 2 = 6 (where m/s can be your unit of measurement).
    Last edited: Apr 28, 2008
  7. Apr 28, 2008 #6


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    Well you could say that the resultant force on the object is given by


    [tex]m\frac{dv}{dt}=F-kv \Rightarrow \frac{1}{F-kv}\frac{dv}{dt}=\frac{1}{m}[/tex]

    and integrating both sides w.r.t. t

    [tex]\int \frac{1}{F-kv}\frac{dv}{dt}dt=\int \frac{1}{m}dt [/tex]

    [tex]\int \frac{1}{F-kv}dv=\int \frac{1}{m}dt [/tex]

    and then go from there. But that is a calculus approach to it.
  8. Apr 28, 2008 #7
    Yeah, and for something simple like that, I would be naturally inclined to do so with calculus. It's just not as obvious for me to use calculus when the expressions are more complicated to integrate (or differentiate).
  9. Apr 28, 2008 #8


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    For full credit, do not use values. Once the equation is solved (in terms of F, k, m), you'll see what general shape it has (i.e, linear, quadratic, exponential, sinusoid, other?)

    You'll need to identify one or several key features of the graph, and label them in terms of the parameters F, k, and m. Without having the actual equation in front of us yet, it's hard to be specific, but features could include:

    The location of any maxima, minima, axis-crossings, and asymptotes.
    If it's a sinusoid: the amplitude, period, and equilibrium value.
    If it's an exponential that decays toward zero: either the time to reach half of the initial value or the time to reach 1/e of the initial value.
    Perhaps the initial slope of the graph at t=0 could be indicated.

    Since you're to graph velocity vs. time, and F/k has the units of a velocity, it may be that v=F/k has some relevance to a feature in your graph.
  10. Apr 28, 2008 #9
    Yeah, I managed to get that far with calculus. I'm looking at the solution to the problem right now. After doing a u-sub for the left side of that, I don't see how they go straight to:

    " ln(F-kv) - ln(c) = (-k/m)t, c being a constant ".

    What did they do with the "+ c's" to get ln(c)?

  11. Apr 28, 2008 #10
    Wait, nevermind. Somehow, it took me a while to see that as a differential equation.
    Last edited: Apr 28, 2008
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