How to Round for Significant Figures in Scientific Calculations?

  • Thread starter sp3sp2sp
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In summary, an elevator ascends 200m with a maximum speed of 5m/s and an acceleration and deceleration of 1.0m/s^2. To find how far the elevator moves while accelerating from rest, the equation (v_f)^2 = (v_i)^2 + 2a_s * delta s is used. However, there may be discrepancies in the rounding of the answer due to the given values being written with different numbers of significant figures. When using Mastering Physics, it is recommended to give answers to 3 significant figures to avoid being marked incorrect.
  • #1
sp3sp2sp
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4

Homework Statement


A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s^2
How far does the elevator move while accelerating to full speed from rest?

Homework Equations



The Attempt at a Solution


delta s = (5m/s)^2 / 1.0m/s^2
= 12.5m
but here's the thing confusing me about rounding.
The values given were written as 5m/s and 1.0m/s^2
When dividing (or multiplying) I thought we use the same number of sig-figs as the value with least amount of sig-figs, which would be "5"
But the answer is 12.5 --three sig-figs. I don't understand why.
Please help thanks
 
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  • #2
sp3sp2sp said:
delta s = (5m/s)^2 / 1.0m/s^2
What equation are you invoking here? If it is 2aΔs = vf2 - vi2, what is Δs when vi = 0?
Aside from that, I think that whoever wrote this question was careless with significant figures. If the answer is given to 3 sig. figs, then all input quantities must also be given to 3 sig. figs.
 
  • #3
sp3sp2sp said:
delta s = (5m/s)^2 / 1.0m/s^2
= 12.5m

You are correct. By the rules of significant figures the answer should be 10.

A little off topic, but I thought I would point out that 5^2/1^2 does not equal 12.5. However, I’m sure that is just a typo. You lost something in the equation.
 
  • #4
Cutter Ketch said:
You are correct. By the rules of significant figures the answer should be 10.
I think you meant to write 1×101. It looks silly, but that's the rule for one sig. fig.
 
  • #5
sp3sp2sp said:

Homework Statement


A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s^2
How far does the elevator move while accelerating to full speed from rest?

Homework Equations



The Attempt at a Solution


delta s = (5m/s)^2 / 1.0m/s^2
= 12.5m
but here's the thing confusing me about rounding.
The values given were written as 5m/s and 1.0m/s^2
When dividing (or multiplying) I thought we use the same number of sig-figs as the value with least amount of sig-figs, which would be "5"
But the answer is 12.5 --three sig-figs. I don't understand why.
Please help thanks
Often an instructor or a textbook will give instructions to treat numbers given in problems as if they have 3 sig-figs, unless otherwise noted in problem instructions.
 
  • #6
its a mastering physics question, so it wouldn't accept my answers until i inputed "12.5"
Im using the third equation of motion: (v_f)^2 = (v_i)^2 + 2a_s * delta s

Thanks for confirming. I am going to say something to prof because I lost points.
 
  • #7
Mastering Physics and other homework delivering platforms expect answers to within 1-3% of the correct answer. Anything beyond that range is marked incorrect. So, although 10 is correct as far as sig figs are concerned, it is marked incorrect because it is 20% off the correct answer. The remedy is, when doing Mastering Physics, to give your answers to 3 sig. figs. when the given numbers have fewer than 3.
 
  • #8
kuruman said:
I think you meant to write 1×101. It looks silly, but that's the rule for one sig. fig.
No, by the rules of significant digits placeholder zeros are not significant. Whichever way you write it that is one significant digit.
 

1. What are significant figures and why are they important?

Significant figures, also known as significant digits, are the numbers that represent the precision of a measurement. They are important because they indicate the accuracy of a measurement and help prevent overestimating the precision of a value.

2. How do you determine the number of significant figures in a given number?

The basic rule for determining significant figures is to count all non-zero digits and any zeros between them. For example, in the number 503.04, there are 5 significant figures (5, 0, 3, 0, and 4).

3. What is the correct way to round a number to the appropriate number of significant figures?

The general rule for rounding is to look at the digit to the right of the last significant figure. If it is 5 or higher, round up the last significant figure. If it is 4 or lower, leave the last significant figure unchanged. However, there are specific rules for rounding in different scenarios, so it is important to consult a guide for specific examples.

4. Can you give an example of rounding for sig-figs?

Sure, let's say we have a measurement of 3.154 cm. The last significant figure is the 4, and the next digit is 5, so we round up the 4 to a 5. The correct rounded measurement would be 3.15 cm.

5. Why is it necessary to round to the appropriate number of significant figures?

Rounding to the appropriate number of significant figures is necessary because it helps maintain the accuracy and precision of a measurement. It ensures that the number of digits used does not overstate the precision of the measurement, which can lead to incorrect calculations and conclusions.

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