Physics lab, calculating/combining errors

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SUMMARY

This discussion focuses on calculating and combining errors in physics lab experiments, specifically addressing the calculations for T^2, d, and a with their respective uncertainties. For T = 0.433 ± 0.004 s, T^2 is calculated using the formula T^2 = T^2 * (2 * ΔT / T). For distance d, given v = 55 mi/hr ± 5% and t = 5.00 ± 0.03 hrs, the formula d = vt is applied, and the uncertainty Δd is derived using Δd = Δ(vt) = vt (ΔV / V + Δt / t). Finally, for acceleration a = 2d/t^2, with d = 82.0 ± 0.2 cm and t = 4.23 ± 0.05 s, the uncertainty Δa is calculated using Δ(2d/t^2) = 2d/t^2 (Δd/d + Δt^2/t^2).

PREREQUISITES
  • Understanding of basic physics concepts, including time, distance, and acceleration.
  • Familiarity with error propagation techniques in measurements.
  • Knowledge of significant figures and their application in calculations.
  • Ability to perform basic arithmetic operations with uncertainties.
NEXT STEPS
  • Study error propagation methods in physics, focusing on formulas for combining uncertainties.
  • Learn about significant figures and their importance in experimental physics.
  • Explore practical applications of kinematics equations in real-world scenarios.
  • Review examples of calculating uncertainties in various physics experiments.
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Students in physics labs, educators teaching experimental physics, and anyone involved in scientific measurement and data analysis.

cjweidle
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I just started the semester off in physics lab and I am so rusty. I did three perfectly okay by myself but these last three I can't seem to figure out. Any help on the direction to take with these is appreciated!

c) T = 0.433 ± 0.004 s. Calculate T^2 andΔT^2.

d) Given d = vt, v = 55 mi/hr ± 5% and t = 5.00± 0.03 hrs, calculate d and Δd.
Like here do I put in d= (55 ± 5%)(5.00 ± 0.03)
I know I have to get the percent into an actually number, I can do that. but I don't know if this is the way to go about it and then I really don't know how to get Δd.

e) Given a = 2d/ t2 , d = 82.0 ± 0.2 cm and t = 4.23 ± 0.05 s, calculate a and Δa.
 
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Hint:

C) If X = a ± b then ΔX ^ n = X ^ n * ( n * ( ΔX / X ) ) = a ^ n ( n * (b / a ) )

D) Δd = Δ(vt) = vt ( ΔV / V + Δt / t )

E) Δ(2d/t^2) = 2d/t^2 ( Δd/d + Δt^2/t^2)
 

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