What is the Maximum Speed of a Porsche 944 Turbo at Full Power?

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The maximum speed of a Porsche 944 Turbo, with a rated engine power of 217 hp and 30% power loss in the drivetrain, was calculated to be 11.34 m/s after considering the effective force from friction. The calculation involved determining that 70% of the engine power translates to approximately 113,317.4 W at the wheels. The total mass of the car and driver is 1530 kg, with two-thirds of the weight over the drive wheels, contributing to the friction force. The user expressed confusion over the low speed result, questioning the role of static friction in achieving maximum velocity. Ultimately, the discussion highlights the complexities of calculating maximum speed due to various forces acting on the vehicle.
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Homework Statement



A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1530 kg, and two-thirds of the weight is over the drive wheels. What is the speed of the Porsche at maximum power output?

Homework Equations


P=FV


The Attempt at a Solution


I calculated the maximum power by taking 70% of 217 and multiplying by 746W/hp = 113317.4W. I don't know where to go from here, since I'm not sure how to obtain a value for F, the forces doing work on the Porsche at maximum velocity. It's not simply rolling friction, since plugging in (2/3)(1530)(9.8)(.02) (.02 is the value of rolling friction listed in the text) for F yields a maximum speed of 566.8 m/s... I would think that there would also be a drag force doing work at maximum velocity, but the dimensions of a cross section of the car are not given. Any ideas on how to go about finding vmax?
 
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Well, I found the correct answer...but it doesn't make any sense to me. Using the friction force that accelerates the car as F (2/3)(1530)(9.8)=9996N, I found v to be 11.34m/s. That seems awfully slow for a sports car and for a car at maximum velocity I didn't think that static friction would be doing work.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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