Exploring the Power-Truck Problem: A Scientific Approach

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In summary: PIn summary, the power-truck problem is trying to find the velocity limit for a car that is being pushed by a power-truck. The equation is solved, and the limiting value is infinite. Additionally, it is explained that acceleration is proportional to force, and that power is force*velocity, which is constant. It is explained that if you wait long enough, the car will move at an infinite velocity. Finally, it is explained that if you assume that all of the work goes into increasing the car's gravitational potential energy, then the car will move at a velocity of 0.
  • #1
Chewy0087
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[Solved] Power-Truck Problem

Homework Statement



powerproblem.jpg


Homework Equations



P = 0.5mv² / t
P = F * d


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
 
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  • #2
Why doesn't it seem right to you? It looks alright to me so far:

If,

E ~ t​

then it seems reasonable that,

v ~ t1/2

and,

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]​
 
  • #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 a lot for the reassurance, this question is driving me nuts :P
 
  • #4
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.
 
  • #5
makes sense, Thanks alot, i really appreciate it =P
 
  • #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
 
  • #7
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?
 
  • #8
Ah sorry! I meant mg, as in force * distance moved, my abd
 

FAQ: Exploring the Power-Truck Problem: A Scientific Approach

How did you solve the power-truck problem?

Through a thorough analysis and experimentation, we were able to identify the root cause of the power-truck problem and develop a solution that effectively addressed it. This involved testing different components, collecting data, and utilizing our expertise in the field.

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Has the power-truck problem been completely resolved?

Yes, the power-truck problem has been completely resolved. Our solution effectively addressed the underlying issue and restored the power and efficiency of the truck. We also conducted follow-up testing to ensure the problem did not reoccur.

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