How do I find the instant when two moving objects are together using a graph? How do I find the instant when two moving objects are together using a graph?
The problem is as follows:

The graph from below describes the motion of an electric car and a
  diesel truck. It is known that they pass through the same point at
  $t=0$. Find the instant when they are together again.


The alternatives given on my book are as follows:
$\begin{array}{ll}
1.&\textrm{12 s}\\
2.&\textrm{15 s}\\
3.&\textrm{18 s}\\
4.&\textrm{16 s}\\
5.&\textrm{20 s}
\end{array}$
For this problem the only thing that I could come up with was to identify the two equations for speed for both the truck and the electric car which are as follows. I'm using the labels $\textrm{t=truck and c=electric car}$
$v_{t}(t)=10$
$v_{c}(t)=2t$
from equating both I obtainted the time when they do have the same speed. (For brevity purposes I'm omitting the units but they are consistent)
$2t=10$
$t=5$
But from then on I don't know what to do with that information or how do I join it with other equation to get to the time when they are together.
Can somebody help me with this?.
 A: The distances traveled are the areas under the lines. Thus, for the electric car, the distance traveled up to $6$ seconds, using the $\frac{bh}{2}$ formula for right-angled triangle areas, is $\frac{12(6)}{2} = 36$. Thus, at any time $t \gt 6$, the total distance traveled would be $36 + 12(t - 6)$.
For the diesel truck, the distance traveled up to time $t$ would be $10t$.
Since the distance traveled by the diesel truck at $t = 6$ is $60 \gt 36$, it's ahead of the electric car. As such, since the speed of the electric car is $12$ which is greater than the diesel truck's speed of $10$, there will be a time $t \gt 6$ when the $2$ vehicles are together again given by
$$36 + 12(t - 6) = 10t \tag{1}\label{eq1A}$$
I trust you can finish the rest yourself.
A: Let's think of it like this. Let $c(t),d(t)$ denote the positions of the car and diesel truck respectively. Given their grounding in reality, we may assume these functions are continuous and, given the graphs provided, differentiable.
We know that $c(0) - d(0) = 0$ since they are at the same position at $t=0$. (Equivalently, $c(0) = d(0)$.)
At some time $t = \tau > 0$, $c(\tau) = d(\tau)$, similarly. We wish to figure out this $\tau$.
Note that these together imply $c(\tau) - c(0) = d(\tau) - d(0)$. This bears resemblence to the fundamental theorem of calculus. The derivatives $c'(t),d'(t)$ represent the velocity of each vehicle. Accordingly, the above relation implies, through that theorem,
$$\int_0^\tau c'(t)dt = \int_0^\tau d'(t)dt$$
Recall the analogy of areas to integrals: $\int_a^b f(t)dt$ is the signed area under $f(t)$ on the interval $[a,b]$.
Thus, given this fact and the graph you've been shown, we want to find $t=\tau>0$ such that the areas under each velocity function are equal. 
You can also approach this analytically instead, with $c'(t) = 2t$ for $0 \le t \le 6$ and $c'(t) = 12$ for $t > 6$, and $d'(t) = 10$.. Then you want $\tau>0$ such that the above integral equality is satisfied. Using the property of $\int_a^b f(t)dt + \int_b^c f(t)dt = \int_a^c f(t)dt$, and the definitions ofr $c',d'$, we can then see
$$\underbrace{\int_0^6 2t \; dt + \int_6^\tau 12 \; dt}_{\displaystyle{ \int_0^\tau c'(t)dt}} = \underbrace{\int_0^\tau 10 \; dt}_{\displaystyle{\int_0^\tau d'(t)dt}}$$
Of course, you would get rid of the middle integral on the interval $[6,\tau]$ above and make the leftmost integral on $[0,\tau]$ if it turned out that $\tau < 6$, but the area analogy makes it very clear $\tau > 6$ if you use the provided graphs.
In any event, all that remains is to find the antiderivatives, apply the fundamental theorem of calculus, and solve for $\tau$!
