Significant figures in one of the Sears and Zemansky book problems?

• FountainDew
In summary, the conversation discusses the calculation of instantaneous acceleration at t=1.0s using the given formula. Rounding off to significant figures can lead to loss of important information. The solution converges to a=1.0m/s^2 as delta t approaches smaller values.
FountainDew

Homework Statement

Average acceleration:
- The instantaneous velocity at any time can be calculated with the given formula: v = 60 m/s + (0.50 m/s^3) t^2
- Find the instantaneous acceleration at t=1.0s, by taking (delta)t to be 0.1s, then 0.01s, and then 0.001s.

Homework Equations

a = (delta)v / (delta)t

The Attempt at a Solution

@t = 1.0s
v = 60.5 m/s

@t=1.1s
v = 60 m/s + (0.50 m/s^3)(1.1 s)^2
v = 60.61 m/s

a = (delta)v / (delta)t = (60.61 m/s - 60.5 m/s) / (0.1 s) = 1.1 m/s^2

etc... (repeated for 0.01s and 0.001s, not important to my question)

4. The actual solution
@t = 1.0s
v = 60.5 m/s

@t=1.1s
v = 60 m/s + (0.50 m/s^3)(1.1 s)^2
v = 60.605 m/s

a = (delta)v / (delta)t = (60.605 m/s - 60.5 m/s) / (0.1 s) = 1.05 m/s^2
etc...

5. The Question
If 1.1 squared is 1.21, then if it is multiplied with 0.50 it should equal 0.605. But according to the rule of multiplication of significant figures, the result is equal to the least amount of sig fig's. So it should equal 0.61? This is significant to me, because as you go down the question, sig figs matter more and more and more!

What am I doing wrong?

I am not sure if you have had calculus yet, but the instantaneous acceleration at t=1.0 seconds is 1.0 m/s^2. As you take your delta t to smaller and smaller values, the solution converges to a = 1.0. If you round off to sig figures, it's like throwing the baby out with the bathwater.

PhanthomJay said:
I am not sure if you have had calculus yet, but the instantaneous acceleration at t=1.0 seconds is 1.0 m/s^2. As you take your delta t to smaller and smaller values, the solution converges to a = 1.0. If you round off to sig figures, it's like throwing the baby out with the bathwater.

Thanks for the quick reply! And I didn't do so well in calculus, but I'm going to keep trying! The problem made more sense once I started keeping the numbers after the decimal intact, I could see it slowly moving to 1.0 but never reaching it. By rounding it early, I lost the entire sense of the problem! The baby and the bathwater analogy helped me understand, thanks for that imagery. :)

1. What are significant figures and why are they important in scientific calculations?

Significant figures are the digits in a number that are known with certainty, plus one digit that is estimated. They are important because they indicate the precision of a measurement or calculation and help ensure the accuracy of scientific data.

2. How do you determine the number of significant figures in a measurement or calculation?

The rules for determining significant figures are:

• All non-zero digits are significant.
• Zeroes between non-zero digits are significant.
• Leading zeroes (to the left of the first non-zero digit) are not significant.
• Trailing zeroes (to the right of the last non-zero digit) may or may not be significant, depending on the measurement or calculation.

3. Can you round significant figures in a calculation?

Yes, when performing calculations, the final answer should be rounded to the same number of significant figures as the measurement with the least number of significant figures. This helps maintain the accuracy of the calculation.

4. How do significant figures affect scientific notation?

When writing a number in scientific notation, the number of significant figures is represented by the number of digits in the coefficient. For example, the number 0.0056 written in scientific notation would be 5.6 x 10^-3, with two significant figures.

5. Why is it important to pay attention to significant figures in scientific calculations?

In science, precision and accuracy are crucial. Paying attention to significant figures ensures that the results of calculations are as precise and accurate as possible. It also helps to avoid misleading or incorrect data in scientific research.

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