How Much Power Does a Top Fuel Drag Racer Generate?

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SUMMARY

A top fuel drag racer with a mass of 500.0 kg completes a quarter-mile (402 m) drag race in 4.3 seconds, achieving a final speed of 115 m/s. The average power output of the engine can be calculated using the formula P = W/T, where W is the work done. The work can be derived from the change in kinetic energy, leading to the conclusion that the average power output is determined by the work-energy theorem, specifically W = ΔE_k. The discussion emphasizes the importance of understanding acceleration and force in the context of variable forces during the race.

PREREQUISITES
  • Understanding of basic physics concepts such as force, work, and energy.
  • Familiarity with kinematic equations and their applications.
  • Knowledge of the work-energy theorem and its implications.
  • Ability to perform calculations involving average power and acceleration.
NEXT STEPS
  • Study the work-energy theorem in detail, focusing on its applications in variable force scenarios.
  • Learn how to derive acceleration using kinematic equations, specifically vf - vo/t.
  • Explore advanced topics in dynamics, including the effects of friction and air resistance on performance.
  • Investigate the relationship between power output and vehicle performance in drag racing contexts.
USEFUL FOR

Physics students, automotive engineers, drag racing enthusiasts, and anyone interested in the mechanics of high-performance vehicles.

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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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