I am modelling the time it will take for a person to slide down a zip line from a fixed start to end point in zip lines of different shapes. The models will be performed with straight line, parabolic curve, elliptical arc and cycloid.
I am able to solve for the parabolic and elliptical curve; however, I am encountering issues while solving the time taken for the parabolic curve. For starters, I derive the equation for time from mechanics as well as conservation of energy. This equation assumes that gravity is the only force that acts on the object and that friction is negligible.
As the start point and end points of my model is fixed, I first find the equation of a parabola that passes through the points A (start) and B (end). Therefore, I am left with this equation of a parabola which is the model for my zip line:
Now, when I try to find the time taken by the passenger using the equation for time which is the first image; an error arrises on my calculator that says 'Non real values'. I do not understand why this error comes. The calculation is shown below:
Why does the error come on the calculator? The exact same formula works without showing the error for the straight line and elliptical arc. What is making it show 'Non real values' for this integral? What can be done in order to attain a value for time as it is crucial for my investigation? I need the travel time in seconds it takes for the passenger on this particular zip line.
From conservation of energy $$v^2=2g(y_1-y)$$ Then your formula becomes $$T=\int_{x_1}^{x_2}\sqrt{\frac{1+\left(\frac{dy}{dt}\right)^2}{2g(y(x_1)-y(x))}}dx$$ The formula is valid only for $y<y(x_1)$, meaning that you can never go higher than the initial point (if you started with zero velocity). You notice then that if $x_1=0$ in $y(x)=0.000174x^2-0.6852$ it means that the object starts at the bottom of a parabola, and it will not be able to go up (no energy). In order for the problem to work, you need to start at $x_1\ne 0$ and have $y(x_2)\le y(x_1)$.