Explain to me Step Functions for VT graphs

Click For Summary

Homework Help Overview

The discussion revolves around understanding step functions in the context of distance-time and velocity-time graphs. The original poster expresses confusion regarding the representation of velocity in the velocity-time graph based on the slopes calculated from the distance-time graph.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the slopes of the distance-time graph to the corresponding sections of the velocity-time graph, questioning the accuracy of their interpretation. Some participants inquire about the visibility of the graphs and the implications of the calculated average velocities.

Discussion Status

The discussion is ongoing, with participants exploring the relationship between average velocity and the shape of the velocity-time graph. There is an indication that the original poster's understanding may need adjustment, particularly regarding the behavior of velocity over time.

Contextual Notes

Participants note the absence of visible graphs, which may hinder the clarity of the discussion. The original poster mentions that both graphs share the same intervals, which could be relevant to the interpretation of the data.

kencamarador
Messages
33
Reaction score
0
I am sort of confuse about this step functions. In the picture below, the top is the distance time graph and the bottom is velocity time graph. The arrow pointing is what is confusing me. Its not suppose to be a step up (answers in my book) but i calculated the slopes

First curve line = 30
Second straight diagonal line = 60
Second curve line = 35

So I put it in my velocity time graph but the part where the arrow is pointing, isn't that correct though?

But the correct answers show that the first line on the velocity time graph is connected to the horizontal second line.

I am so confuse...

Both graphs have same intervals.

http://imgur.com/52DSmKv

http://imgur.com/52DSmKv
 
Physics news on Phys.org
No graphs are visible!
 
The problem is with the first curved line. The average velocity from 0 to 6 is 30, but the velocity is varying from zero at t = 0 to a value higher than 30 at t = 6, in order for the average to be 30. If the velocity is varying linearly with time in this region, then the velocity at t = 6 has to be 60 in order to the average to be 30. Then there won't be any step.
 

Similar threads

Analyse motion of an oscillator: x(t)=0.2cos(12*pi*t)
  • · Replies 5 ·
Replies
5
Views
2K
Inconsistency in a Velocity-Time Graph
  • · Replies 13 ·
Replies
13
Views
4K
Finding acceleration from Velocity vs Position graph
  • · Replies 2 ·
Replies
2
Views
5K
C 12B 2021 #23Analyzing Motion Graphs: Understanding Slopes and Acceleration
  • · Replies 2 ·
Replies
2
Views
2K
Confusion Calculating Young's Modulus
  • · Replies 5 ·
Replies
5
Views
2K
Translating motion graphs. Dt to vt
  • · Replies 2 ·
Replies
2
Views
2K
Motion in one dimension -- Experiments with a Hot Wheels car rolling down a ramp
  • · Replies 12 ·
Replies
12
Views
5K
Understanding a Velocity-Time Graph
  • · Replies 14 ·
Replies
14
Views
6K
Finding the change in velocity from an acceleration vs time graph
  • · Replies 1 ·
Replies
1
Views
3K
What is the intersection point of two objects on a position vs time graph?
  • · Replies 10 ·
Replies
10
Views
3K