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Homework Help: Graphical Relationship Between ∆x and v

  1. Jul 16, 2011 #1
    You're pushing a car up an inclined plane (a hill). Which graph matches the relationship between ∆x and v?

    2. Relevant equations

    W = K.E. + P.E.
    thus: F∆xcos[itex]\theta[/itex] = 1/2mv2 + mgh

    3. The attempt at a solution

    The answer is C. I thought the answer was D - I don't understand why the relationship is a negative. Please share an explanation with me.

    Attached Files:

    Last edited: Jul 16, 2011
  2. jcsd
  3. Jul 16, 2011 #2
    yes answer is c

    Ax= Acos(theeta)

    so (delta)x = 1/2 * Ax*t^2

    and v= A * t

    two equations give

    delta(x) = 1/2 * cos(theta) * v^2 * 1/A

    gives a prabola on x-axis
  4. Jul 16, 2011 #3
    Thanks darkxponent! Could you tell me which component in the equation F∆x = 1/2mv2 I could specifically look at and determine the graph is a negative exponent function? Is it because there's a "1/2"? If that is true... will any coefficient < 1 result in a negative exponential function?
  5. Jul 16, 2011 #4
    its not exponential but a parabolic function

    y=Ax2 is an eqn of parabola with y as its axis.
    more on parabolas: http://en.wikipedia.org/wiki/Parabola
  6. Jul 16, 2011 #5

    by using diffrential calculus u can find the shape of the graph

    if diffrentiation>0 implies graph increeasing
    if diff<0 inmplies graph decreasing

    and if double diff>0 :graph up shaped { y= x^2}
    if double diff<0 :graph down shaped {y = - x^4}

    magnitude of coefficient has nothing to do with the shape of graph
    but + or - sing changes the shape of the graph only to create mirror image of the same
  7. Jul 16, 2011 #6
    Okay, ignoring all else where is the negative component in this equation F∆xcos[itex]\theta[/itex] = 1/2mv^2??

    Here's the practice problem verbatim: A student applies force to a stalled car over a distance ∆x to increase its kinetic energy. Which graph BEST represents the relationship between the car's speed and the pushing distance?

    Please don't use calculus in the explanation - I'll be clueless.
  8. Jul 16, 2011 #7
    u cant even think of physics without calculus!!!
  9. Jul 16, 2011 #8
    Yes you can - just memorize the equations : ) Besides I wasn't an engineering or physics major in college and took physics without calculus as electives.
  10. Jul 19, 2011 #9

    1) The correct graph is a square root function, so it looks like the inverse of an exponential function. ...parabola... hyperbole... sigmoidal... etc... none-of that mattered because I just wanted to relate the mathematical function of the graph to the equation I gave, and not so much the graph's shape and how it moves n such. :smile:

    2) This is a square root function because when you solve for v in the equation F∆xcosθ = 1/2mv2 + mgh ... we get [itex]\sqrt{}[/itex][2(F∆xcosθ - mgh)/m] = v

    So basically I stopped one step ahead of finishing the problem... the function they were asking for (per the graphs) was v(∆x) not v2(∆x)! Otherwise I would have gotten the question correct :redface:

    Thanks for your help!