Concept question on using integrals to find displacement

Click For Summary
SUMMARY

The discussion centers on calculating total displacement using the integral of the velocity function v(t) = 22 + 18t over the interval from t = 1.5 s to t = 3.1 s. The correct approach involves evaluating the definite integral, which requires calculating (3.1^2 - 1.5^2) rather than (3.1 - 1.5)^2. This distinction is crucial for accurately determining displacement, as it reflects the application of the Fundamental Theorem of Calculus in evaluating integrals of polynomial functions.

PREREQUISITES
  • Understanding of calculus concepts, specifically integrals
  • Familiarity with polynomial functions and their properties
  • Knowledge of the Fundamental Theorem of Calculus
  • Ability to evaluate definite integrals
NEXT STEPS
  • Study the Fundamental Theorem of Calculus in detail
  • Practice evaluating definite integrals of polynomial functions
  • Learn how to apply integrals to real-world displacement problems
  • Explore common mistakes in integral calculus and how to avoid them
USEFUL FOR

Students studying calculus, educators teaching integral calculus, and anyone seeking to understand the application of integrals in physics for calculating displacement.

soggybread
Messages
20
Reaction score
0

Homework Statement



I was reading a solution to the following question:

Given v(t) = 22 + 18 t, use calculus to determine the total displacement from t initial= 1.5 s to t final = 3.1 s

Homework Equations



Here are the steps in the equation:
http://qaboard.cramster.com/Answer-Board/Image/20091122250246336739742469787505163.gif

I get everything up to the last part, where we take get (3.1^2 - 1.5^2)... Why is that we don't do (3.1-1.5)^2 instead? I had always imagined the it would be [tex]\Delta[/tex]t squared, so just final minus initlal, then squaring the result, and not the final time squared minus initial time squared.

Would be great if you guys could clear up my confusion.
 
Last edited by a moderator:
Physics news on Phys.org
Then you imagined wrong. (3.1^2 - 1.5^2) is right, (3.1-1.5)^2 is wrong. If you've done a lot of integrals with a lower limit of 0, you might have gotten that feeling. But it's not correct.
 

Similar threads

Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K