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

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)

[Dyuman Joshi]

I believe there's a typo in the problem statment. It should be 0-60m/s

[Dyuman Joshi]

@Sam, what is the relation between velocity and time? Do you have an expression/equation?

Answer 1

[Voss]

% 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')

[Dyuman Joshi]

@Voss, Shouldn't we not use a linear interpolation for a non-linear data/function?

[Sam]

Worked perfect thanks mate.

[Voss]

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]

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