Maximum acceleration of a Porsche on a concrete surface?

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Homework Help Overview

The discussion revolves around calculating the maximum acceleration of a Porsche 944 Turbo on a concrete surface, considering its power output, mass, and the static friction coefficient. Participants are exploring the relationship between power, force, and acceleration in the context of vehicle dynamics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of maximum acceleration using static friction and the normal force. Questions arise regarding the correct application of forces and the impact of weight distribution on the drive wheels.

Discussion Status

The discussion is active, with participants offering insights into the forces at play and questioning the assumptions made about the normal force. Some guidance has been provided regarding the calculation of static friction and its role in determining acceleration.

Contextual Notes

Participants are considering the effects of weight distribution on the drive wheels and whether to ignore air resistance in their calculations. There is an emphasis on understanding the relationship between power, force, and acceleration without reaching a definitive conclusion.

philo51
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A Porsche 944 Turbo has a rated power of 217 hp and a mass, with the driver, of 1460 kg. Two-thirds of the car's weight is over the drive wheels.

What is the maximum acceleration of the Porsche on a concrete surface where static friction=1.0

1hp=476watts
217hp=103292 watts

and power= Force*velo but what's the force?

What is the speed of the Porsche at maximum power output
thats easy if i get the force i just solve for v

If the Porsche accelerates at , how long does it take until it reaches the maximum power output?
again easy if i can get the first part. any help is appreciated
 
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The force acting on the car is the static friction of the road on the tires. The maximum possible static friction force is given by [itex]\mu N[/itex], where N is the normal force pressing the surfaces together. That should get you going.
 
yea i think i got that but I am not sure if the rest is right then

u*N=ma

(1.0*1460kg*9.81)/1460 = a? that doesn't seem right is there another force?
 
Friction is the only horizontal force you need worry about. (I presume you may ignore air resistance.) But note that you were told "Two-thirds of the car's weight is over the drive wheels". How would that modify the normal force used to calculate the friction?
 
ooo right so

1*9.81*(2/3)m=m(total)*a ok that should workk thanks!
 

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