Concept question on using integrals to find displacement

In summary, the question asks to use calculus to determine the total displacement of an object from a given initial time to a final time, using the equation v(t) = 22 + 18t. The solution involves taking the integral of the velocity equation and plugging in the final and initial times. The correct expression is (3.1^2 - 1.5^2), not (3.1-1.5)^2 as the latter would be incorrect.
  • #1
soggybread
20
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
  • #2
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.
 

Related to Concept question on using integrals to find displacement

1. How do you use integrals to find displacement?

To use integrals to find displacement, you first need to have a function that represents the velocity of an object over a given time interval. Then, you can use the integral of this function over the same time interval to find the total displacement of the object.

2. What is the difference between displacement and distance?

Displacement refers to the change in an object's position from its starting point to its ending point, while distance refers to the total amount of ground covered by the object. Displacement takes into account the direction of motion, while distance does not.

3. Can integrals be used to find displacement in multiple dimensions?

Yes, integrals can be used to find displacement in multiple dimensions. This is done by integrating the velocity function with respect to time in each dimension separately.

4. Are there any limitations to using integrals to find displacement?

One limitation is that the velocity function must be continuous and differentiable over the given time interval. Additionally, this method may not accurately capture changes in velocity that occur rapidly or irregularly.

5. How is the area under a velocity-time graph related to displacement?

The area under a velocity-time graph represents the displacement of an object. This is because the integral of the velocity function over a given time interval gives the total change in position, which is equivalent to displacement.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
393
  • Calculus and Beyond Homework Help
Replies
3
Views
629
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
2
Views
914
  • Introductory Physics Homework Help
Replies
6
Views
794
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
259
Back
Top