A Porsche 944 Turbo has a rated engine power of 217 hp (161882 W). 70% of the power is lost in the drive train, and 30% reaches the wheels. The total mass of the car & driver is 1480 kg, and two thirds of the weight is over the drive wheels. What is the max acceleration if the coefficient of static friction is 1? (2/3)(1480 kg)(9.80)*1= 9669.33 N Force of static friction Set this = to mtotal* a 9669.33N=(1480)a a= 6.53 m/s^2 What is the speed of the Porsche at max power output? P=Fv F= 9669.33 N P= (161882 W) * (.70) = 113317.4 W 113317.4= (9669.33 N)v v=11.72 m/s which is about 26.2 mph Does this seem fast enough???
The calculation of acceleration is correct, but the second question is not well defined. The speed you calculated is the maximum speed when the acceleration is still limited by friction and not by engine power, but even this is true only if one gear is set to maximize power at that speed (the engine's power depends on rpm).
You original statement needs to be revised to read 30% loss in drive train, 70% reaches the wheels. Your calculations seem correct, assuming the car is geared so that max power occurs at an acceleration rate of 1 g. However, the second question isn't worded to state that the same gearing used for max acceleration of 1g would be used for determining the car's speed at max power. It would have been better stated to ask for the maximum speed that the maximum acceleration of 1g occurs at. 30% seems a little high for drive train losses, generally it's usually around 15%, although an all wheel drive car could have higher drive train losses.
so much so wrong thats a problem with numbers out side the real world a 26.2 mph sports car ??????? a porsche 944 is far closer to the idea 50/50 weight front rear power loss is about 15% and they top 140 mph