Can You Calculate Acceleration Using Power or Work?

[timman_24]

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?

 

[Nenad]

$$F = ma$$

$$W = Fd$$

$$P = \frac{W}{t}$$

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.

 

[timman_24]

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...

 

[krab]

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.

 

[timman_24]

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

 

[krab]

accel is dv/dt. Look again at the formula I gave. It is essentially ma=P/v-kv^2