Calculating Velocity and Acceleration for Variable Motion

Click For Summary

Homework Help Overview

The discussion revolves around calculating velocity and acceleration for an object in motion, described by the position function X = 20t - 5t^3. Participants are exploring the conditions under which velocity and acceleration are zero, as well as calculating average velocity over a specified time interval.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to derive expressions for velocity and acceleration from the given position function. Questions arise regarding the calculations of average velocity and the interpretation of results, particularly concerning the values obtained for time and velocity.

Discussion Status

Some participants have provided calculations and results, while others are seeking clarification on specific values. There appears to be a mix of interpretations regarding average velocity, with some guidance offered on the definition and calculation methods. The discussion is ongoing with no explicit consensus reached.

Contextual Notes

Participants are working under the constraints of the problem as presented, with no additional information provided. The original poster expresses uncertainty about their results, indicating a desire for verification against others' calculations.

monkfishkev
Messages
11
Reaction score
0
An object starts from position X = 0 and moves along a straight line with its position given by; X = 20t - 5t^3 where t is the time taken to reach position X.

Find;
i) The time at which the velocity is zero
ii) The time at which the acceleration is zero
iii) The average velocity over the first three seconds of motion

No answers were given with the questions, so would appreciate some answers to check against mine. I have 1.15s, 0s and -50m/s
 
Physics news on Phys.org
yes,yes, no(-25)
 
Can you please explain how you come to get -25. Thanks for your time :-)
 
I must be doing this wrong:

v = 20 - 15t^2
After 1s, v = 20 - 15(1) = 5m/s
After 2s, v = 20 - 15(4) = -40m/s
After 3s, v = 20 - 15(9) = -115m/s

Average speed = 5 - 40 - 115 =
= -150 ÷ 3 = -50m/s
 
You are looking for average velocity. This is defined as displacement / time.
 
avg. velocity=(∫vdt)/(∫dt),but here ∫vdt is x itself(given) so for three seconds
x=-75
t=3
<v>=-25
 
thank you
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
886
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
12
Views
2K