# Need help on how to find the uncertainty in this question

## Homework Statement

To find the acceleration of a glider moving down a sloping air track, you measure its
velocities (v1 and v2) at two points and the time t it takes between them, as follows:
v1 = 0.210 +/- 0.05m/s
v2 = 0.850 +/- 0.05m/s
t = 8 +/- 1 second
What should you report for the acceleration, a = (v2 - v1)/ t and its uncertainty

a = (v2 * v1)/ t

## The Attempt at a Solution

Acceleration, a = (V2 – V1) / t = ((0.85 – 0.21) / 8.0) m/s = 0.08 m/s2

I need some clue on how to find the uncertainty in this question

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## Answers and Replies

Bystander
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t = 8 +/- 1 second

??? Is that a clue?

Bystander
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"The uncertainty in the time measurement is ± 1s."

gneill
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You need some more Relevant equations, namely the error (uncertainty) propagation formulas for combining the uncertainties when you add, multiply, and divide values with uncertainties.

Take a look http://www.rit.edu/cos/uphysics/uncertainties/Uncertaintiespart2.html [Broken]

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But how do I calculate the uncertainty using t = +/- 1 or is it the answer to the uncertainty?

mfb
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It is something you will need to calculate the uncertainty.

You can start with a part that might be easier: what is the uncertainty on the velocity difference v2-v1?

gneill
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But how do I calculate the uncertainty using t = +/- 1 or is it the answer to the uncertainty?
All the measured values (velocities, times) have uncertainties associated with them. +/- 1s is the uncertainty in the time measurement of 8 seconds. So the "actual value" could be as low as 7 seconds or as high as 9 seconds. You need to combine the uncertainties through your calculations following the established rules for doing so. So, do the subtraction v2 - v1 first and establish the uncertainty in that intermediate value. Then move on to the division operation.

Example: x = (2.0 ± 0.2) cm, y = (3.0 ± 0.6) cm. Find z = x - 2y and its uncertainty.

z = x - 2y = 2.0 - 2(3.0) = -4.0 cm
Dz = Dx + 2 Dy = 0.2 + 1.2 = 1.4 cm

So z = (-4.0 ± 1.4) cm.
Using Eq 1b, z = (-4.0 ± 0.9) cm.

The 0 after the decimal point in 4.0 is significant and must be written in the answer. The uncertainty in this case starts with a 1 and is kept to two significant figures. (More on rounding in Section 7.)

I tried to relate my problem to the given example but in my case ,
v1 = 0.210 +/- 0.05m/s
v2 = 0.850 +/- 0.05m/s

delta A = (0.05 - 0.05 )/ 0.1 = 0 ???

I am lost here. Please help

haruspex
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z = x - 2y = 2.0 - 2(3.0) = -4.0 cm
Dz = Dx + 2 Dy = 0.2 + 1.2 = 1.4 cm
Note that although the expression for z is x - 2y, Dz is given as Dx+2Dy. This is because it is calculating the extreme possible values of z. The lowest possible value is when x takes its lowest value but y is at its highest value.
You can extend this method for all of your uncertainties. For each variable, which extreme value produces the lowest possible acceleration and which the highest possible?

That said, there is another way such questions can be interpreted: statistically. This takes into account that not all variables are likely to be at the extremities of their error ranges at the same time. But based on your prior example, that's not what you are expected to do here.

May I know how do I go about to find the uncertainty for my question since I was told to find the v2 -v1 first then move on to division operation

haruspex
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May I know how do I go about to find the uncertainty for my question since I was told to find the v2 -v1 first then move on to division operation
As I said, investigate this:
For each variable (v1, v2, t), which extreme value produces the lowest possible acceleration and which the highest possible?
(Technically, we should not assume that the lowest possible acceleration as v1 varies will correspond to an extreme value of v1, etc. But it will be the case here.)

Hi
I have calculate the uncertainty but I am not really sure whether it is correct.
Uncertainty = 0.007(1 Sig fig)
Hence, A= 0.08 +/- 0.007 m/s^2

Could anyone helps me check if my answer is correct?

lightgrav
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how did you get that .007 m/s2 ?
how to get a small acceleration:
what's the smallest change the velocity might have made? (0.54 m/s) divide it by the longest time possible
Largest a is just the opposite.

Using the uncertainty formula I got 0.007 but I am not really if I apply it correctly
The book answer is 0.08 +/- 0.01

I have made a error in my question . The time t should be 8 +/- 0.1 second

haruspex
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Using the uncertainty formula
What uncertainty formula? Are you referring to something you already posted in this thread? If so, I don't seem to be able to identify it.
The book answer is 0.08 +/- 0.01
That's not the answer I get.

Given that v1 is in the range 0.210- 0.05m/s to 0.210+ 0.05m/s, and v2 is in the range 0.850 - 0.05m/s to 0.850 + 0.05m/s, what is the maximum possible value of v2-v1?

I have made a error in my question . The time t should be 8 +/- 0.1 second
Ah, that makes a difference.... but I still don't get the book answer. Well, to two significant figures I do, so guess that's ok.

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Hi

I have solved this question with the advise given by all of you :)