Graphing Terminal Velocity using Euler's Method

Click For Summary

Homework Help Overview

The discussion revolves around calculating terminal velocity for an object shot upward, considering factors such as mass, drag coefficient, and gravitational acceleration. The original poster is using Euler's Method to compute values and graph acceleration versus time, while seeking feedback on the accuracy of their graphing results.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster shares their code and results, expressing uncertainty about the graph of acceleration versus time and its correlation with the velocity graph. Participants discuss the shape of the curves and the presence of a "wiggle" in the acceleration graph, questioning its implications on the velocity graph.

Discussion Status

Participants are actively engaging with the original poster's concerns, providing observations about the graphs and discussing the relationship between acceleration and velocity. There is no explicit consensus, but some participants affirm the shape of the curves while exploring the reasons behind the observed "wiggle."

Contextual Notes

There is mention of a specific drag coefficient and the effects of changing it on the graphs. The original poster's professor has provided feedback on the velocity graph, which influences the ongoing discussion about the acceleration graph's behavior.

jinksys
Messages
122
Reaction score
0
Suppose an object of 20kg is shot upward with an initial velocity of 20m/s. The drag coefficient I've chosen is 0.25, and gravity is 9.8m/s. I'm trying to calculate the terminal velocity using Euler's Method (using a C prog), and then graph the data using openoffice. I know I am converging to the correct terminal velocity, however I'm not confident in my graph of acceleration vs time.

Does this look right?

http://i36.tinypic.com/10p4jmh.png"

Here is my code:
PHP:
#include <stdio.h>
#include <unistd.h>
#include <math.h>
#include <stdlib.h>

double accel(double v);
double velocity(double v, double a, double t);


int main()
{
double a,v,t,time;
v=0.0;
t=0.01;
a=1.0;
time=0.0;
printf("Velocity, Acceleration, Time,\n");

while(fabs(a)>0.00001)
	{
	a=accel(v);
	printf("%f, %f, %f,\n",v,a,time);
	v=velocity(v,a,t);
	time+=0.01;
	}

}

double accel(double v)
{

double a,k,g,m;
k=0.25;
g=9.8;
m=20.0;

if(v!=0.0){
	a=(-k*v*v*fabs(v));
	a/=m*v;
	a-=g;
	return a;
	}

	return -9.8;
}

double velocity(double v, double a, double t)
{
v = v + (a * t);
return v;
}
 
Last edited by a moderator:
Physics news on Phys.org
Hi jinksys,

The shape of the curve looks fine to me; was there something in particular that bothered you about it?
 
alphysicist said:
Hi jinksys,

The shape of the curve looks fine to me; was there something in particular that bothered you about it?

Well, my professor said that my graph of velocity vs time http://i38.tinypic.com/51368i.png" is correct. However, since my acceleration is essentially a graph of the velocity's slope, the "wiggle" around -9.8 should represent something on my velocity graph around 0m/s. He's not seeing where I get the wiggle.

So you're saying the graph looks correct? What is causing the wiggle?
 
Last edited by a moderator:
jinksys said:
Well, my professor said that my graph of velocity vs time http://i38.tinypic.com/51368i.png" is correct. However, since my acceleration is essentially a graph of the velocity's slope, the "wiggle" around -9.8 should represent something on my velocity graph around 0m/s. He's not seeing where I get the wiggle.

So you're saying the graph looks correct? What is causing the wiggle?


The wiggle on the a vs. t graph occurs at arount t=2, and that corresponds to what is happening at t=2 on the v vs. t graph.

If you look at the acceleration curve, it starts out at about -14.8 or so, and then the wiggle is at -9.8m/s, and then it rapidly goes to zero.

This corresponds to the velocity curve: initially it has a slope of about -14.8; the slope changes to -9.8 around t=2 seconds and the wiggle is showing that that slope stays approximately constant arount t=2 seconds.


I think if you change your drag coefficient to about 2.5, and look at the a and v curve at around t=1 second, it is easier to see what is going on. (Notice that around t=0 the v curve is sloped, around t=1 is approximately straight, and past t=1 it curves again.)



The thing to remember is that the acceleration curve gives the slope of the velocity curve: so these wiggles (which are where the acceleration curves are not changing very much) indicates that the velocity slope is roughly a straight line at that time.
 
Last edited by a moderator:

Similar threads

  • · Replies 2 ·
Replies
2
Views
981
Replies
23
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K