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I'm given the following problem:
At what times is the particle found at $x= 20m$?
I know this is a very fundamental problem, but still I cannot see how the answer ends up being $4$ and $12$. Note that I'm asked to use "motion diagrams" and/or vector arithmetic, not any formulae here.
The key idea is that the total displacement of the object is the area under the velocity-time graph. If you understand this, the rest is trivial.
Calculate the area under the graph at different instants of time, and find the time(s) at which the area is 20. (Since you're working in units where $v_x$ is measured in m/s, and $t$ in seconds, this corresponds to $x= 20$ m.)
If you're confused as to why there are two times that are possible, keep in mind that that the area under the time axis is considered to be negative. You should be able to see that the total area first increases to $40$ m, and then decreases, so it should pass $x=20$ m at two different times.
Use the rules of geometry for the area of a triangle and the area of trapezoid. The displacement is always the area under the curve of velocity with time.
For the first stage, the area is a triangle with $t$ as base and $v_x$ for height
For the second stage, the area is a composite shape, and you can work it out based on the function that describes $v_x$ as a function of time.
