What is the average acceleration of a cart with photogates 30 cm apart?

[curiouspup34]

If the distance between the photogates in the diagram is 30 cm, and the times for the photogates on the LabQuest are listed as:
Blocked 0.00000
Unblocked 0.00156
Blocked 0.035
Calculate the average acceleration (in fundamental-standard units) of the cart. You should assume and initial velocity of 0 m/s and round your answer to 1 decimal place.

I know the average acceleration equation is delta v over delta t and I know how to find the change in velocity but I'm confused on what times to use.

My attempt:
Velocity=0.30m/(0.035)=8.5714m/s
Acceleration= (8.5714m/s-0m/s)/(0.035s-0s)=244.9m/s^2

 

[haruspex]

Velocity=0.30m/(0.035)=8.5714m/s

Obviously, the velocity is not constant. Be specific about what velocity this is.

 

[curiouspup34]

Obviously, the velocity is not constant. Be specific about what velocity this is.

This would be the final velocity

 

[haruspex]

This would be the final velocity

Not according to the way you calculated it.

 

[curiouspup34]

Not according to the way you calculated it.

Then the average velocity? If this is so, is my issue that I need to find the final velocity because they gave me the initial already?

 

[haruspex]

Then the average velocity? If this is so, is my issue that I need to find the final velocity because they gave me the initial already?

Yes.
You will have to assume constant acceleration (which makes the request to find the average acceleration a bit misleading).

 

[curiouspup34]

Yes.
You will have to assume constant acceleration (which makes the request to find the average acceleration a bit misleading).

So I would have
v final = (0.30m/.035s)(2) = 17.143 m/s
Avg acceleration = (17.143m/s)/(0.035s)=489.8 m/s^2

 

[haruspex]

So I would have
v final = (0.30m/.035s)(2) = 17.143 m/s
Avg acceleration = (17.143m/s)/(0.035s)=489.8 m/s^2

Yes, but as I wrote you have to assume constant acceleration in order to claim that final velocity, so it is misleading to call the answer an "average" acceleration. If the acceleration is not constant then the average acceleration will not have that value.

 

[curiouspup34]

Yes, but as I wrote you have to assume constant acceleration in order to claim that final velocity, so it is misleading to call the answer an "average" acceleration. If the acceleration is not constant then the average acceleration will not have that value.

Yeah that's true! Hm okay well I will assume constant acceleration now and go from there. Thank you for your help!