Part I. Vo= 0.4318 meters V= 1.245 meters T= 3.54 seconds Uncertainty= + 0.11 Ho

Question

Part I.

Vo= 0.4318 meters

V= 1.245 meters

T= 3.54 seconds

Uncertainty= + 0.11

How do I calcuate the acceleration INCLUDING the uncertainty?  

Part II.

I am writing a conclusion for a Physics lab.  I am using a manual stopwatch to measure the amount of time (in seconds) for a photogate to cover a certain distance on an air track. We measured it about 20 times, and got an average of about 3.54 seconds.  

a).  briefly discuss the ACCURACY of the manual time measurements???

b).  briefly discuss the precision and uncertainty of the calculation??

if you could give me some ideas, thatd be great.   

Explanation / Answer

part 1.

Let U(y) represent the uncertainty of "y" and y is a function of x1, x2, ..., x_i.

So for your case, a is a function of x and t. i.e. a(x, t)

Then the uncertainty of y is given by:

U(y) = sqrt[ sum{ (c_i)^2 * (x_i)^2 } ]

where c_i = d{y} / d{x_i}
(i.e. derivative of y with respect to x_i)

So the uncertainty of a is found by:

U(a) = sqrt[ ( d{a}/d{t} )^2 * (uncertainty in t)^2 + ( d{a}/d{x} )^2 * (uncertainty in x) ^2 ]

d{a}/d{t} = -4x / (t^3) = -4x / ( 3375) = -0.0012 * x
d{a}/d{x} = 2 / (t^2) = 2 / (225) = 0.0089
uncertainty in t = 2

U(a) = sqrt[ (-0.0012 * x)^2 * 2^2 + (0.0089) ^2 * (uncertainity in x) ^2 ]

Plug in your value for x and your uncertainty for x into the equation above to find the uncertainty in acceleration.

If you are going to present this in a report, include the calculation from above I have just showed you. After the calculation, show your final result:

" a = 786

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