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Homework Help: Power-Truck Problem

  1. Jul 14, 2009 #1
    [Solved] Power-Truck Problem

    1. The problem statement, all variables and given/known data


    2. Relevant equations

    P = 0.5mv² / t
    P = F * d

    3. The attempt at a solution

    Hey sorry again :P trying to sort out some problems...
    For part 1, I'm not sure, basically, I did the equation;

    P = 0.5mv^2 / t
    Re-arranging to Pt = 0.5mv^2

    Now velocity as a function of time;

    [tex]\sqrt{\frac{2Pt}{m}}[/tex] = v (or d(displacement)/dt)

    So displacement = [tex]\int\sqrt{\frac{2Pt}{m}}[/tex]
    = [tex]\frac{t^1.5\sqrt{2P}}{{root of M}}[/tex]

    Doesn't seem right. And;

    Acceleration = [tex]\frac{1}{2}[/tex][tex]\sqrt{\frac{2P}{mt}}[/tex]

    Sorry I know it's long but any help = great :P from here I think I can tackle the rest of the question.. also excuse my numerous latex fails :P
    Last edited: Jul 14, 2009
  2. jcsd
  3. Jul 14, 2009 #2


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    Why doesn't it seem right to you? It looks alright to me so far:


    E ~ t​

    then it seems reasonable that,

    v ~ t1/2


    x ~ t3/2

    However, I think you might be missing some constants from your integral:

    [tex] \sqrt{\frac{2P}{m}} \int t^{\frac{1}{2}} dt = \sqrt{\frac{2P}{m}} \left(\frac{2}{3}t^{\frac{3}{2}}\right) + v_0 [/tex]​
  4. Jul 14, 2009 #3
    Hmm I see, however, the problem then arises for the next question;

    find the limiting value for velocity;

    Using my answer the limiting value would be + infinite.

    It would however make sense I suppose for part C where the acceleration would be infinite as t --> 0 & 0 as t --> infinite. Would that suffice do you think?

    Thanks alot for the reassurance, this question is driving me nuts :P
  5. Jul 14, 2009 #4


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    Does this really surprise you? If you supply something steady power, P, forever, then its energy will continue to increase without bound.

    Acceleration is proportional to force. Power is force*velocity, which is constant. The longer you wait, the larger the velocity is, which means that the amount of force available to you is much lower. That's your argument for t --> infinity.

    Reversing this reasoning gives you an argument for t --> 0.
  6. Jul 14, 2009 #5
    makes sense, Thanks alot, i really appreciate it =P
  7. Jul 16, 2009 #6
    Me again :Z just trying to get some closure on this ;X

    concerning e) & f);

    Would saying P=[tex]\frac{10g * d}{t}[/tex]

    Going to [tex]\frac{Pt}{10g}[/tex] = Distance Moved. Be correct?

    Then differentiating with respect to time --> [tex]\frac{P}{10m}[/tex] = v

    Be correct? :P

    Then for f) I don't understand the question, sureley it's 1:1 assuming no resistance? All of the KE --> potential energy? Or am i missing something?

    No idea but maybe;

    P = [tex]\frac{10g*d}{t}[/tex] + 0.5mv^2

    & when v is very small? :[, no idea.

    Thanks again for any help
  8. Jul 16, 2009 #7


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    I'm a little confused about what the number '10' is in your solution to (e). Have you considered taking the power to be the rate at which work is done, and then assuming that all of the work goes into increasing the car's gravitational potential energy?
  9. Jul 16, 2009 #8
    Ah sorry! I meant mg, as in force * distance moved, my abd
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