Instantaneous Velocity at t=4: Solving Displacement Equation s= t^2- 5t + 15

Click For Summary

Homework Help Overview

The problem involves finding the instantaneous velocity from a displacement equation given by s = t^2 - 5t + 15 at t = 4. The context is rooted in calculus, specifically in the application of limits and derivatives to determine velocity from a position function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of instantaneous velocity and the relevance of limits in its calculation. There are inquiries about the original poster's understanding of derivatives and the formula provided. Some participants encourage the original poster to share their attempted methods for better insight into their confusion.

Discussion Status

The discussion is ongoing, with participants providing guidance on the interpretation of the displacement function and the calculation of instantaneous velocity. There is an emphasis on understanding the underlying concepts rather than simply applying formulas. Multiple interpretations of the problem are being explored, and no consensus has been reached yet.

Contextual Notes

There is a suggestion that the original poster may not have fully grasped the concept of derivatives or the significance of the limit process in calculating instantaneous velocity. The discussion hints at a potential gap in foundational knowledge regarding the relationship between displacement, average velocity, and instantaneous velocity.

Amber430
Messages
15
Reaction score
0
Q: Displacement in a straight line is s= t^2- 5t + 15. What's the instantaneous velocity when t= 4?


Equation: lim [f(a+h)-f(a)]/h as h approaches 0

I tried this problem and what I came up with for an idea of how to solve this makes absolutely no sense.
 
Physics news on Phys.org
Have you covered rules for derivatives yet?
 
Even if you think your method of solution makes absolutely no sense you should still tell us what you did. This way we can see what the real problem is.

Do you understand what the formula you were given represents? Normally when you're given a distance function and you want to calculate the average velocity between two time intervals you will calculate s2-s1 divided by the time it took. The smaller you take this "width" of time the closer you get to the instantaneous velocity. Does it make sense now?
 
To expand just a bit on what Cyosis wrote, and looking at your displacement as a function of t, you have s(t) = t^2- 5t + 15.

What you need to do is to calculate
[tex]\lim_{h \rightarrow 0} \frac{s(4 + h) - s(4)}{h}[/tex]
That will be your instantaneous velocity at t = 4.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
9K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K