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## Main Question or Discussion Point

A position of a particle in linear motion is given by:

x = vt + 0.5at

Calculate x with the error for:

t = 25.3 ± 0.5s

v = 10.1 ± 0.4m/s

a = 2.5 ± 0.3m/s

So for calculating vt:

q = (10.1) (25.3) = 255.53 (exact)

Δq = (10.1)(25.3) * √ (0.4/10.1)

= 11.31...

Therefore, vt = 255.53 ± 11.3 (the 11.3 is rounded because I heard uncertainties always rounded to precision of last uncertainty)

Now my question is:

How do you round 255.53? And how many sig figs?

Is it 260.0 ± 11.3?

Or 255.5 ± 11.3?

OR something else?

And why?

I thought 260.0

Because when there is just a number to a certain position with no uncertainty given, like 123.4 it is implied that:

123.4 ± 0.1

So the number is rounded to the lowest precision of the uncertainty (I think)

(i.e. if the uncertainty was ± 0.12, the rounding would still occur at the 10th decimal place.

In this case the tenth decimal place.

In the case of 255.53 ± 11.3, it is the 10's.

Therefore 260.0

I'd appreciate if someone would type of a full solution (doesn't have to be pretty).

I guess a similar question is lets say you have the number

1234.5678 +/- 123.45

How should 1234.5678 be rounded?

x = vt + 0.5at

^{2}Calculate x with the error for:

t = 25.3 ± 0.5s

v = 10.1 ± 0.4m/s

a = 2.5 ± 0.3m/s

^{2}So for calculating vt:

q = (10.1) (25.3) = 255.53 (exact)

Δq = (10.1)(25.3) * √ (0.4/10.1)

^{2}(0.5/25.3)^{2}= 11.31...

Therefore, vt = 255.53 ± 11.3 (the 11.3 is rounded because I heard uncertainties always rounded to precision of last uncertainty)

Now my question is:

How do you round 255.53? And how many sig figs?

Is it 260.0 ± 11.3?

Or 255.5 ± 11.3?

OR something else?

And why?

I thought 260.0

Because when there is just a number to a certain position with no uncertainty given, like 123.4 it is implied that:

123.4 ± 0.1

So the number is rounded to the lowest precision of the uncertainty (I think)

(i.e. if the uncertainty was ± 0.12, the rounding would still occur at the 10th decimal place.

In this case the tenth decimal place.

In the case of 255.53 ± 11.3, it is the 10's.

Therefore 260.0

I'd appreciate if someone would type of a full solution (doesn't have to be pretty).

I guess a similar question is lets say you have the number

1234.5678 +/- 123.45

How should 1234.5678 be rounded?