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Homework Help: Graphing Terminal Velocity using Euler's Method

  1. Oct 15, 2008 #1
    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?


    Here is my code:

    #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;
    printf("Velocity, Acceleration, Time,\n");

        printf("%f, %f, %f,\n",v,a,time);


    double accel(double v)

    double a,k,g,m;

        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. jcsd
  3. Oct 16, 2008 #2


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    Homework Helper

    Hi jinksys,

    The shape of the curve looks fine to me; was there something in particular that bothered you about it?
  4. Oct 16, 2008 #3
    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
  5. Oct 16, 2008 #4


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    Homework Helper

    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
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