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

  • Thread starter Thread starter scytherz
  • Start date Start date
  • Tags Tags
    Automobile Power
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
2 replies · 3K views
scytherz
Messages
10
Reaction score
0

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??
 

Discussion

Physics news on Phys.org
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...