How Does Friction Affect a Car's Maximum Acceleration?

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Friction plays a crucial role in determining a car's maximum acceleration, particularly for a rear-wheel-drive vehicle that distributes 42% of its weight on the drive wheels with a static friction coefficient of 0.80. The maximum acceleration can be calculated using the formula F_net = ma, where F_net represents the frictional force acting on the rear tires. To find this force, one must first determine the normal force on the rear tires and then apply the static friction coefficient. Once the maximum acceleration is established, calculating the time to reach 105 km/h becomes straightforward. Understanding these principles is essential for solving the problem effectively.
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Homework Statement



A rear-wheel-drive car supports 42% of its weight on its two drive wheels and has a coefficient of static friction of 0.80.

What is the car's maximum acceleration?

Assuming infinite engine power, how long will it take the car to reach 105 km/h?

Homework Equations



F_g = mg

µ_s(F_n) = F_f

The Attempt at a Solution



Unsure how to begin problem, D;.
 
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What is u_s(F_n) on the rear tires? That's the friction force that accelerates the car forward per Newton's 2nd law. What is the car's acceleration, and then find the time it takes to reach the given speed?
 
Ya I know what I'm supposed to find, it says that in the problem. To be honest the second part of the problem is easy once the first part is done. I just don't know how to go about starting the first part.
 
use Newton 2, F_net =ma, where F_net is the maximum available horizontal force acting forward on the rear tires. What is that force?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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