Can You Calculate Acceleration Using Power or Work?

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Acceleration can be calculated using power and work by manipulating the fundamental equations of physics. The relationship between force, power, and velocity is established with the formula F = P/v, allowing for the integration of acceleration into the equations. The discussion highlights the need to account for air resistance, which is proportional to the square of velocity, complicating the calculations. A derived equation, m(dv/dt) = P/v - kv^2, illustrates how to express acceleration in terms of power while factoring in drag. Overall, the challenge lies in accurately incorporating these variables to derive a usable formula for acceleration from power.
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I know acceleration can be found by the following basic equation:
A=F/M

But my question is can you also find acceleration using power or work?

Background:
I am doing a project in physics right now that involves taking the torque, weight, and drag of the car and calculating the theoretical 0-60, 1/4, and 1/8 mile times. The problem is cars are rated in horsepower (power) and torque (work.) I don't exactly know how to 1) Break them down to their force properties and solve or 2) make an equation that will solve for acceleration.

Can anyone help out?
 
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F = ma

W = Fd

P = \frac{W}{t}

sub in Work into Power

W = Fd

P = \frac{Fd}{t}

since V = \frac{d}{t}

P = FV

play around with those, see what you get.
 
Well I messed around and got these... But it still isn't what I am looking for. Anyone else have some input?


P=FxTxA
P=pxA
P=MxVxA

I subed in momentum for FxT.
Then I subed in Mass and Velocity for Momentum...

Doesn't help me much, but it does get acceleration in the equation...
 
Nenad told you what how to get force from power: F=P/v. Use F=ma=dv/dt. You still need air resistance losses, and these are proportional to v^2.
m{dv\over dt}={P\over v}-kv^2
This gives v vs. t and you need to integrate once more to get distance vs. t.
You can get the proportionality constant k knowing the top speed and power, or the drag coefficient.

BTW, torque is not work.
BTW, given an adequate transmission, torque is a spurious performance parameter.
 
Okay I ended up getting power from velocity and air resistance using the drag formula:

P=(cD)(A)(d)(v)^3

Now what I need is not getting power from force but getting the accel from power. Anyone have any suggestions?

Thanks
 
Last edited:
accel is dv/dt. Look again at the formula I gave. It is essentially ma=P/v-kv^2
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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