$1.2.14.$ According to the graph of acceleration versus time, set the speed at times $4$ and $15$ s, if at time $1$ s the speed is $3$ $\frac{m}{s}$.
Acceleration depends on time, as: $$\left\{\begin{matrix} a(t) = 0\;m/s^2,\; 0\text{ s} \leq t \leq 2\text{ s}\\ a(t) = 20t,\; 2\text{ s} \leq t \leq 5\text{ s}\\ a(t) = 60\;m/s^2,\; 5\text{ s} \leq t \leq 9\text{ s}\\ a(t) = 60-20(t-9),\; 9\text{ s} \leq t \leq 12\text{ s}\\ a(t) = 0, t> 12\text{ s} \end{matrix}\right.$$ Given that the area under the graph of acceleration vs. time is velocity vs. time. Then the velocity depends on time as: $$\left\{\begin{matrix} v(t) = 3\;m/s,\; 0\text{ s} \leq t \leq 2\text{ s}\\ v(t) = 3+20t^2/2,\; 2\text{ s} \leq t \leq 5\text{ s}\\ v(t) = 93+60(t-5),\; 5\text{ s} \leq t \leq 9\text{ s}\\ v(t) = 333 - 20 (t - 9)^2/2 + 60 (t - 9),\; 9\text{ s} \leq t \leq 12\text{ s}\\ v(t) = 423\;m/s, t> 12\text{ s} \end{matrix}\right.$$ The graph of this dependence is presented below:
It follows that at $4$ and $15\text{ s}$ the velocity is $43$ and $423\;m/s$, respectively.
$$v_1=43\;m/s; \;v_2=423\;m/s.$$