# Homework Help: Graphing Terminal Velocity using Euler's Method

1. Oct 15, 2008

### jinksys

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: Apr 23, 2017
2. Oct 16, 2008

### alphysicist

Hi jinksys,

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

3. Oct 16, 2008

### jinksys

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: Apr 23, 2017
4. Oct 16, 2008

### alphysicist

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: Apr 23, 2017