Displacement, velocity, acceleration

Click For Summary

Homework Help Overview

The discussion revolves around a particle's motion described by the position function s(t) = t^3 - 8t + 1, focusing on displacement, average and instantaneous velocity, acceleration, and direction changes over a specified time interval.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore calculations for displacement, average velocity, instantaneous velocity, and acceleration at specific times. There are attempts to verify answers and clarify the average velocity calculation.

Discussion Status

Some participants express uncertainty about specific answers, particularly regarding average velocity. There is a recognition of potential errors in initial assumptions, and guidance is sought to clarify the calculations.

Contextual Notes

Participants are checking their answers against the problem requirements, indicating a focus on accuracy and understanding of the concepts involved in motion analysis.

Jacobpm64
Messages
235
Reaction score
0
A particle moves along a line so that its position at any time t>0 is given by the funciton s(t) = t^3 - 8t + 1, where s is measured in feet and t is measured in seconds.
(a) Find the displacement during the first 3 seconds.
(b) Find the average velocity during the first 3 seconds.
(c) Find the instantaneous velocity when t = 3.
(d) Find the acceleration of the particle when t = 3.
(e) At what value or values of t does the particle change direction?

Ok, this is pretty much just to check my answers because I'm paranoid.. Here are the answers i came out with..

(a) 4 ft
(b) 1 ft/s
(c) 19 ft/s
(d) 18 ft/s^2
(e) sqrt(8/3)s or 1.633333s

Is this correct?

Thanks
 
Physics news on Phys.org
Jacobpm64 said:
(b) 1 ft/s
Check this one again. Other than that the answers seem fine.
 
LeonhardEuler said:
Check this one again. Other than that the answers seem fine.

Yea except for that one I think they're all right.
 
let's see.. for that one i did..

s(t) = t^3 - 8t + 1
s(0) = 0^3 - 8(0) + 1 = 1
s(3) = 3^3 - 8(3) + 1 = 4
so it would be
average velocity = change in displacement/change in time
average velocity = (4 - 1) / (3 - 0) = 3/3 = 1 ft/s

not sure on this? can you point me in the right direction?
 
Jacobpm64 said:
let's see.. for that one i did..
s(t) = t^3 - 8t + 1
s(0) = 0^3 - 8(0) + 1 = 1
s(3) = 3^3 - 8(3) + 1 = 4
so it would be
average velocity = change in displacement/change in time
average velocity = (4 - 1) / (3 - 0) = 3/3 = 1 ft/s
not sure on this? can you point me in the right direction?

Ohh yea I guess you're right I just wasn't thinking right and thought that initial displacement was zero, you're doing it right.
 
Uhh, yeah never mind-oops!
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
Replies
2
Views
2K
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K