how do I find the x-axis value from my yline intercept on my acceleration curve

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Sam
Sam 2023-11-1
评论： Dyuman Joshi 2023-11-3
I have an acceleration curve and i need to find the 0-60mph (0-26.8224m/s) Stuck on how to get there, currently only have a yline that shows where 26.8224m/s along the y-axis. Please help! Here's my code and a picture of the graph:
figure(6)
P1 = plot(tout,yout(:,1));
xlabel('Time (s)')
ylabel('Velocity (m/s)')
grid on
set(P1,'linewidth',2)
title('Acceleration Curve')
yline(26.8224)

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采纳的回答

Voss
Voss 2023-11-1
Ran in:
% some data:
tout = (0:60).';
yout = 10*log(tout+1);
y_val = 26.8224;
% your plot:
P1 = plot(tout,yout(:,1),'linewidth',2);
xlabel('Time (s)')
ylabel('Velocity (m/s)')
grid on
title('Acceleration Curve')
yline(y_val)
% find the t value where y is y_val:
t0 = interp1(yout(:,1),tout,y_val)
t0 = 13.6259
% plot a vertical red line to that point on the curve:
hold on
plot([t0,t0],[0,y_val],'--r')
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显示 2更早的评论
Voss
Voss 2023-11-3
Linear interpolation gives a point on the curve MATLAB plotted. If there's a non-linear function underlying that curve, use an interpolation method suitable to that function in your judgment.
Dyuman Joshi
Dyuman Joshi 2023-11-3
Ran in:
But you chose a non-linear function, then why did you use linear interpolation?
Also, the point obtained by linear interpolation is not correct -
syms t
eqn = 10*log(t+1) - 26.8224 == 0;
t = solve(eqn, t)
t = 
vpa(t)
ans = 
13.617800523282271447111158159562

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