How Fast Can a 1967 Corvette Go Using Full Power in 6 Seconds?

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SUMMARY

The discussion focuses on calculating the speed of a 1967 Corvette with a 427 cu-in engine rated at 435 hp after 6 seconds of full power acceleration. The correct speed attained is 120 mph, contrary to initial incorrect calculations of 8.99 mph. Key equations used include the power equation P = W/t and the kinetic energy formula ke = 0.5mv^2. The participants emphasize that the engine's work should be considered in terms of kinetic energy rather than gravitational force.

PREREQUISITES
  • Understanding of basic physics concepts such as power, work, and kinetic energy.
  • Familiarity with the equations P = W/t and ke = 0.5mv^2.
  • Knowledge of horsepower conversion to watts (1 hp = 746 W).
  • Ability to perform unit conversions, particularly from meters per second to miles per hour.
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  • Learn about the efficiency of internal combustion engines and its impact on performance.
  • Explore advanced vehicle dynamics and acceleration calculations.
  • Investigate the effects of weight and engine specifications on vehicle speed and acceleration.
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scytherz
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Homework Statement



A 1967 Corvette has a weight of 3020 lbs. The 427 cu-in engine was rated at 435 hp at 5400 rpm. a) If the engine used all 435 hp at 100% efficiency during acceleration, what speed would the car attain after 6 seconds? b) What is the average acceleration? (in “g”s).


Homework Equations



P= W/t
ke =.5mv^2
1 hp = 746 W = 550 ft lb/s

The Attempt at a Solution



P= (fd)/t , since f = Mg, P= MgV,
P= 435 hp = 324 510 kg m^2/s^2 (W) = Mgv = 1372.73 kg(9.8m/s^2)(v)
i got v= 24/6 = 4.02m/s.. converting to mph, ans is 8.99 mph... which is wrong.. the ans.. is 120mph...



I also tried using P = W/t , by finding the work, w= pt, w= 1 947 060 Nm, then substitute on k= mv^2, but still wrong...

I don't where did i go wrong, but do i still to put into consideration the 427 cu in engine and the efficiency rate or just ignore it??
 
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scytherz said:

The Attempt at a Solution



P= (fd)/t , since f = Mg, P= MgV,
P= 435 hp = 324 510 kg m^2/s^2 (W) = Mgv = 1372.73 kg(9.8m/s^2)(v)
i got v= 24/6 = 4.02m/s.. converting to mph, ans is 8.99 mph... which is wrong.. the ans.. is 120mph...



I also tried using P = W/t , by finding the work, w= pt, w= 1 947 060 Nm, then substitute on k= mv^2, but still wrong...

I don't where did i go wrong, but do i still to put into consideration the 427 cu in engine and the efficiency rate or just ignore it??

Your first attempt makes no sense at all. The engine does no work against gravity, but
all the work goes into increasing the kinetic energy. f is not equal to Mg.
Your second attempt should work, but the kinetic energy is (1/2)mv^2
 
willem2 said:
Your first attempt makes no sense at all. The engine does no work against gravity, but
all the work goes into increasing the kinetic energy. f is not equal to Mg.
Your second attempt should work, but the kinetic energy is (1/2)mv^2

Ive tried that.. but since the 427 in cu engine contains fuel to run the car, i assumed that this chemical energy will transform to kinetic energy when moving (since burning 1 L of gasoline is equal to 3.5 x10^7 j/s , so I substituted it in the formula, and still got a wrong answer... does my thoughts of using the fuel as the kinetic energy wrong? if so, what would be the best way to approach in this problem...
 

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