How Much Power Does a Top Fuel Drag Racer Generate?

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Homework Help Overview

The discussion revolves around calculating the average power output of a top fuel drag racer, given its mass, distance traveled, and time taken to complete a quarter-mile drag race. The problem involves concepts from kinematics and energy, specifically focusing on the relationships between force, work, and power.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various methods to determine acceleration and force, questioning the assumptions of constant acceleration. Some suggest using the work-energy theorem, while others discuss the implications of different approaches to calculating power.

Discussion Status

The discussion is active, with participants offering different perspectives on how to approach the problem. Some have provided insights into using energy concepts, while others are clarifying misunderstandings regarding the calculations of acceleration and average velocity.

Contextual Notes

There are indications of confusion regarding the use of kinematic equations and the assumptions of constant acceleration. Participants are also navigating the implications of using different methods to calculate work and power.

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Homework Statement


A top fuel drag racer with a mass of 500.0 kg completes a quarter-mile (402 m) drag race in a time of 4.3 s starting from rest. The car's final speed is 115 m/s. What is the engine's average power output? Neglect friction and air resistance.


Homework Equations


P = W/T


The Attempt at a Solution


P = W/T
W=FD
F=MA =500*26.744 = 13372
..plugged in for W/T ... (13372 * 402 /4.3 )/ 1000
This has not gotten me far!
Please help and explain! THANKS!
 
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Using a kinematic equation in the first 4.3 seconds it travels 402m starting from rest, so what is its acceleration? When you get this, what is the force exerted by the car then?
 
Yes, how did you get that 26.744 m/s^2? I get two different answers for the acceleration depending on whether I use the distance of 402 m in 4.3 s or the final velocity of 115 m/s in 4.3 s. I think that means the acceleration is not constant, and so you can't do the problem using the W = FD method (because the force isn't constant either).

I suggest you start with the P = W/t but instead of W = Fd, use the final energy minus the initial energy for the W.
 
I got acceleration by vf-v0/2.
However I got the answer. I did P=FV for avg velocity.
Thanks though!
 
I think a better approach here would be to use the work-energy theorem:

W=\Delta E_k
 
RoyalCat,
I cannot see the formula u posted. please try it again
 
I got acceleration by vf-v0/2.
That doesn't make sense; it gives an average velocity but not an acceleration. Wrong units for one thing.

RoyalCat's formula says W = final energy - initial energy
and is the same thing I suggested earlier.
 
Sorry, my typo, I meant to say i got avg velocity by using the above equation. I got acceleration by using vf - vo/t. Using that info, i plugged it into P = FV (avg V)
 

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