$v = -0.01t^3 + 0.22t^2 - 0.4t$. Find the 2 positive values of t for which the particle is instantaneously at rest. This part has been done: $t= 2$ and $t= 20$.
Find the distance travelled by the particle while its velocity is positive.
We find an expression for the distance $s$ and we just substitute $t=2$ and $t=20$ in it and then substract them together. My question is:
Why do we take these values of t?
Here is a plot of your given velocity function. Notice that the velocity is positive positive between $t=2$ and $t=20$, and so when you find the expression by integrating the velocity function, these two points of time become your limits of the integral. (Integral of the velocity is the displacement. In this case, since velocity is always positive, we are only travelling in one direction, so distance = displacement!)
Just to make this clear, $$s = \int_2^{20} v(t) dt = \int_2^{20} (-0.01t^3 + 0.22t^2 - 0.4t) dt = [f(t)]_{2}^{20} = f(20) - f(2)$$
where $f = -0.0025 t^4 + 0.0733333 t^3 - 0.2 t^2$