Average Acceleration of a Accel vs Speed Graph

lboucher

Hi All

So I have an Accel vs Speed graph and data.
Goes from 0 to 80 mph.
This is all the info I have.

I would like to figure out what Average Acceleration would give me the same distance, of an acceleration from 0 MPH to 80 MPH.
AKA, using the formula Distance = (Vfinale^2)/(2*Acceleration)

Any ideas on how to figure this out?

mfb

If you plot 1/acceleration versus speed, your function value is proportional to the time spent at a specific velocity (assuming your acceleration is always positive).
If you plot speed/acceleration versus speed, the area below the function is proportional to the distance travelled.

lboucher

Sorry, but that still confuses me as the units don't add up. Also when I attempt to plot this out in excel and add up the area under the curve, the resulting numbers cannot be right

sophiecentaur

The area under the v/t graph should give total distance, whatever the instantaneous velocity and accelerations are. Does that hold for an (a/v)/v graph too? Is there not a constant needed after the integration?

mfb

A simple example: v(t)=at+c with constant a and c.
Distance after time T is ##\frac{1}{2}aT^2 + cT##.

v/a = t + c/a
t=(v-c)/a
Therefore, v/a = (v-c)/a + c/a = v/a (trivial in this example, as a is constant)
Speed increases from c to c+aT. If we integrate v/a from c to c+aT, we get ##\int \frac{v}{a}dv = \frac{1}{2a}((c+aT)^2-c^2) = cT+\frac{1}{2}aT^2## - the same as above.

sophiecentaur

Wine and gin make this less accessible but the sums seem right in this case. It that also OK for non constant acceleration? I guess it could be - piecewise. Good one. I love it when someone else does the algebra and I can follow it.

Chestermiller

Not just proportional to the distance traveled. It is equal to the distance traveled. Another way to do this is to plot 1/(2a) as a function of v². This will also be equal to the distance traveled.

mfb

I did not check the prefactor when I wrote my first post here, but it is 1, right.

It works there as well, but the integration can get more difficult.