Ed asked in 社會與文化語言 · 10 years ago

MATLAB 問題 (會英文的)

作業問題不會做

#1:

A parachutist of mass m=68.1kg jumps out of a stationary hot air balloon. The velocity of the parachutist as a function of time is given by the equation: v(t)=gm/c(1-exp(-(c/m)t))

(a) Plot the exact solution of the position as a function of time by integration of v(t) for 0 <=t<=30 with an increment of 1 s. On the same graph, plot the position found by employing the trapezoidal rule to numerically integrate v(t) using an increment of 1 s. Include a legend.

(b) On a second graph, plot the position as a function of time for 0<=t<=30 with an increment of 1 s using the MATLAB quad function. Compare the results to the exact solution. Include a legend.

Try the MATLAB subplot command to place the graphs of parts (a) and (b) in the same graphics window.

會的幫一下!

Update:

我不是要翻譯

我要會MATLAB的幫我做這一題

Update 2:

The drag coefficient c is 12.5 kg/s and g is the gravitational constant g=9.81 m/s^2

Update 3:

2 Answers

Rating
  • Elisha
    Lv 6
    10 years ago
    Favorite Answer

    看到公式後, 發現公式中的c不知為何? 是一參數嗎? 是否可以給個數值呢?以方便計算

    2011-05-16 07:55:34 補充:

    clear all

    clc

    m = 68.1; % kg

    c = 12.5; % kg/s

    g = 9.81; % m/s^2

    % (a)

    tspan = 0:1:30;

    velocity = g*m/c*(1-exp(-(c/m)*tspan));

    % integration by trapezoidal rule

    position_t(1) = 0; % assume the height equal to zero at initial point

    for k = 1:length(tspan)-1

    position_t(k+1) = trapz([velocity(k) velocity(k+1)]);

    end

    % plot

    figure(1)

    plot(tspan, velocity, tspan,position_t)

    xlabel('time,sec')

    ylabel('velocity,m/s')

    legend('velocity','position by trapezoidal rule','location','northwest')

    % (b)

    position_q(1) = 0;

    for k = 1:length(tspan)-1

    position_q(k+1) = quad(@(t) g*m/c*(1-exp(-(c/m)*t)),tspan(k),tspan(k+1));

    end

    % plot

    figure(2)

    plot(tspan, velocity, tspan,position_q)

    xlabel('time,sec')

    ylabel('velocity,m/s')

    legend('velocity','position by matlab of quad rule','location','northwest')% subplot

    figure(3)

    subplot(2,1,1)

    plot(tspan, velocity, tspan,position_t)

    legend('velocity','position by trapezoidal rule','location','southeast')

    subplot(2,1,2)

    plot(tspan, velocity, tspan,position_q)

    legend('velocity','position by matlab of quad rule','location','southeast')

    xlabel('time,sec')

    ylabel('velocity,m/s')

    ----------------------------------------------------圖形如下, 我列最後一張

    圖片參考:http://imgcld.yimg.com/8/n/AF03297167/o/1611051406...

  • 10 years ago

    試著幫你前半部,剩的你試著自己翻看看(掙扎過才會有進步)

    跳傘員質量(即體重)68.1 公斤,從一靜止的熱氣球跳下,該員降落速度與時間的函數關係如公式所述 v(t)=gm/c(1-exp(-(c/m)t))。

    (a) 對 v(t) 積分,範圍 0 <=t<=30,以 1 秒(s) 遞增,繪出時間與位置函數關係的正確圖形。在相同的圖上,再利用梯形積分法,對 v(t) 進行數值積分,遞增值同為 1 秒,將求出的位置點連成圖線。請在圖上加上註解文字。

    2011-05-15 08:49:33 補充:

    這可把我難倒了,我不是理工科的,加上沒學過微積分。

    有請此方面的高手岀馬!

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