Power from mass and acceleration

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LaCalia
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Hello everyone, first post! I'm not terribly fluent in these matters, I'll say to start, but I have a problem.

If a 3,219lb car is capable of accelerating at 40.2ft/s^2, or 0-60mph in 2.19 seconds

How much power must it be capable of producing?

Thanks for your time and patience! :)
 
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There's not enough information. What you have is the average acceleration. What's missing is the maximum acceleration possible due to traction limits of the tires.

Technically it takes zero power for the initial acceleration from zero speed. During the period of maximum acceleration, the associated power will increase linearly with speed, since power = force x speed, and force = mass x acceleration.
 
Not sure I know what to take away from your answer, but I do wish to clarify that 40.2ft/s^2 is the maximum acceleration do to the bounds of the tires.

I calculated the 0-60 time based on that.
 
OK, so maximum acceleration = average acceleration = 40.2 ft / sec^2, due to traction limits. Then the maximum power required is the power required to accelerate a 3,219lb car at 40.2 ft / sec^2 at 60 mph. Note 1 pound mass = 1/32.174 slug (unit of mass). 1 mile = 5280 feet. 1 hour = 3600 seconds. 1 horsepower = 550 ft lb / sec. Force = mass x acceleration.

power = force x speed = ((3219/32.174) x 40.2 ) x (60 x 5280 / 3600) (1/550) ~= 643.5 rwhp (rear wheel horsepower).

It traction allowed for a greater amount of initial acceleration, the required power would be less. The math would be more complicated.
 
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Having a little trouble following your maths, but otherwise, I am eternally grateful for your answer!

Thank you!
 
Ah! Well that helps a ton! A million more thanks!
 
I won't bother doing the math here, but for a similar problem where the traction is limited at the start (called stage 1 in the post linked below), then acceleration limited by power / speed (called stage 2 in the post linked to below). Given a velocity, total time, maximum acceleration, mass, ..., power could be determined using the formula for velocity (v = ... ) from the post linked to below:

https://www.physicsforums.com/threa...-power-requirement.864834/page-2#post-5443890