This is more an AP Physics problem of understanding a graph of velocity, v, as a function of time, t. It's important to note that v = dx/dt (or first derivative of distance x w.r.t. t), so x(t) is areas under the curve (or a definite integral of v w.r.t. t with lower limit of 0). Also, acceleration a = dv/dt (or second derivative of x w.r.t. t) is just derivative of v w.r.t. t.
Since v is continuous line segments here, areas under the curve are easily found by inspection as follows:
1) 0 < t <= 4 : 16 so x(4) = 16 and dv/dt = 0
2) 4 < t <= 7 : 6 so x(7) = 22 and dv/dt = -4/3
3) 7 < t <= 10: -6 so x(10) = 16 and dv/dt = -4/3
4) 10 < t <= 13: -9 so x(13) = 7 and dv/dt = 3/2
5) 13 < t : 7- 2(x-13) assuming v = -2 remains constant for t > 15, and dv/dt = 0.
Note in addition to the obvious areas under curve, I also included each interval's slope dv/dt or acceleration because it's efficient to organize all important values at once so we don't have to waste time figuring out these values that are needed later. Now we can answer all the questions:
A) particle moves forward whenever v > 0. Thus, 0 < t < 7
B) particle moves backward whenever v < 0. Thus 7 < t
C) particle turns around when v changes sign. Thus t = 7
D) particle speeds up whenever dv/dt > 0. Thus, t = 10 even though particle continues its backward motion with less speed after t = 10.
E) Similarly, particle slows down whenever dv/dt < 0. Thus, t = 4 , 13 are 2 points where particle suddenly slows down though particle continues its forward motion at t = 4 until t = 7, and particle cnntinues its backward motion after t = 13.
F) Since each of 5 intervals are linear, acceleration = 0 for all except at t = 4, 10, 13
Since dv/dt = 0 except at x=4, 7, 10, 13 (where dv/dt are undefined), there isn't any particle speed up
G) and H) v_max = 4 for 0 < t < 4. Normally, v_max is found from solving t from dx/dt=0 and dv/dt < 0, but here we have continuous linear segments of v(t), so we just read off the max velocity from graph.
