# Correct rounding for sig-figs

## 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?

## The Attempt at a Solution

delta s = (5m/s)^2 / 1.0m/s^2
= 12.5m
but heres 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 dont understand why.
Please help thanks

## Answers and Replies

kuruman
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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.

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.

kuruman
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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.

SammyS
Staff Emeritus
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## 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?

## The Attempt at a Solution

delta s = (5m/s)^2 / 1.0m/s^2
= 12.5m
but heres 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 dont 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.

its a mastering physics question, so it wouldnt 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. Im going to say something to prof cuz I lost points.

kuruman
Science Advisor
Homework Helper
Gold Member
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.

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.