This should be pretty simple but I'm blanking on this. I need to model (graph) how path 1 becomes equally as efficient as path 2 as the distance of path 2 increases.
distance of path 1 (from A to B) = $10$
distance of path 2 (from A to C) = $y$
The path is a path taken by a forklift driver. The vehicle speed is $2.2\, m/s$. The paths are measured in meters.
On path $1$, there is an added flat time of $30$ seconds (due to something he has to stop and do). On path $2$, there is an added flat time of $10$ seconds.
So in summary:
$$\frac{x}{2.2} + 30 = t_\text{path $1$} \ \ \ \ \ where \ \ \ x>0$$ $$\frac{y}{2.2} + 10 = t_\text{path $2$} \ \ \ \ \ where \ \ \ y>0$$
I know that once the distance of path $2 = 54$ meters, both routes take the same amount of time ($34.545$ seconds). But I can't figure out how to put this into equations that I can use to graph.
I can clarify anything if needed.
Color scheme: $\color{blue}{time}$, $\color{red}{distance}$, $\color{purple}{velocity}$
Given a velocity $v$, the time to go from A to B is $$ \color{blue}{t_{1}}(\color{purple}{v}) = \frac{\color{red}{10}}{\color{purple}{v}} + \color{blue}{30}. $$ The time to go from A to C is $$ \color{blue}{t_{2}}(\color{purple}{v}) = \frac{\color{red}{y}}{\color{purple}{v}} + \color{blue}{10} $$
When $\color{red}{y}$ is small $\color{blue}{t_{2}}<\color{blue}{t_{1}}$; when $\color{red}{y}$ is larger, $\color{blue}{t_{2}}>\color{blue}{t_{1}}$.
Find the value of $\color{red}{y}$ for which $$ \color{blue}{t_{1}}(\color{purple}{v}) = \color{blue}{t_{2}}(\color{purple}{v}) \tag{1} $$
Solving $(1)$, $$ \begin{align} % \color{blue}{t_{1}}(\color{purple}{v}) &= \color{blue}{t_{2}}(\color{purple}{v}) \\ % \frac{\color{red}{10}}{\color{purple}{v}} + \color{blue}{30} &= \frac{\color{red}{y}}{\color{purple}{v}} + \color{blue}{10} % \end{align} $$ The distance is $$ \color{red}{y} = \color{red}{10} + \color{blue}{20}\color{purple}{v} $$
Given $\color{purple}{v}=2.2$, the distance is $\color{red}{y}=\color{red}{54}$.