Distance from time and velocity vectors..HELP

Question

EEEmatlab 2017-11-26

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/369346-distance-from-time-and-velocity-vectors-help

回答： kowshik Thopalli 2017-11-26

so i have 2 vectors:

time = [0,10,15,20,30,59,62,125];
velocity = [0,56.40,97.23,136.25,226.16,403.86,440.44,1265.23];

how would i get distance for each time interval? i tried using

distance = trapz(velocityChange,timeChange);

like this:

time = [0,10,15,20,30,59,62,125];
velocity = [0,56.40,97.23,136.25,226.16,403.86,440.44,1265.23];
distance = zeros(size(time)-1);
for i=1:length(time)
  for ii=1:length(velocity) 
    for x=1:1:7
      velocityChange = velocity(1,(x+1))-velocity(1,x);
      timeChange = time(1,(x+1))-time(1,x);
      distance(x) = trapz(velocityChange,timeChange)
      %%distance = velocity*time%%    
    end 
  end
end

i want distance as a matrix but i get

distance = [ 0 0 0 0 0 00 ] basically a 0 matrix

and i get the overall distance of all the points how do i get the individual distance for each time interval?

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Answer 1

kowshik Thopalli 2017-11-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/369346-distance-from-time-and-velocity-vectors-help#answer_293238

在 MATLAB Online 中打开

trapz

cannot be used here because you are passing only one value for x and y. Moreover direct integration is not the right thing because acceleration is not constant. Instead, what you can do is fit a line (linear regression) between time and velocity.The slope of the line is the acceleration. You can then use d= 0.5*acceleration*((delta_time)^2) to get the distance traveled here for each time interval.

    close all;
scatter(time,velocity);
grid on; hold on;
b1=time'\velocity';% linear regression;
%b1=acceleration
vcalc1=b1*time;
plot(time,vcalc1,'r-');grid on;
 %%to get distance between each time interval
 time_diff=diff(time); % diff- successive differentiation. No need of loop
d= 0.5*a.*(time_diff.^2);
%%output
  d =
       449.28       112.32       112.32       449.28       3778.4       40.435        17832