## Average Acceleration of a Accel vs Speed Graph

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?
 Mentor 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.
 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

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## Average Acceleration of a Accel vs Speed Graph

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?
 Mentor 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.
 Recognitions: Gold Member Science Advisor 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.

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 Quote by 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.
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.

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

 It that also OK for non constant acceleration?
It works there as well, but the integration can get more difficult.