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

AI Thread Summary
The discussion centers on calculating instantaneous acceleration using a given velocity formula and the significance of significant figures in the results. The user initially struggles with the calculations and the implications of rounding, particularly when determining the value of acceleration as delta t decreases. It is clarified that the instantaneous acceleration converges to 1.0 m/s² as delta t approaches zero, emphasizing that premature rounding can obscure the true behavior of the function. The analogy of "throwing the baby out with the bathwater" illustrates the importance of retaining precision in calculations. Ultimately, the user gains clarity on the problem by maintaining decimal accuracy throughout the process.
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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?
 
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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. :)
 
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