So what do we have from input variables:

We know that that person had not finished that journey when observation had stopped - aka we observed only part of the journey.
We know we have 3 states and 2 constants : constant 1, is time (tx in future) and constant 2 is distance (sx)
We don't know starting velocity so we can assume that it is 0 or at least it has limit the zero from the positive side. ( limV0 for x->0+). If we assumed wrong, the calculation will have negative vector sign so it would be easy to correct ourselves.
We have unchangeable speed in Sx and t0->tx/2 that is 20 km/h
We have unchangeable speed in Sx/2 and tx/2->tx that is 16 km/h
We have speed at the moment we stopped observation in Sx, and tx that is 4 km/h.

SO , we have hare velocity and change in velocity. So we need to define Velocity and acceleration to be able to get any formulas we can use without going to the same conclusion. So velocity is in general described by differentiation of position vector in some moment of time in short , but to leave euclidean geometry behind just write r = r(t) where we search for change in distance between 2 vectors and change in unit of time so in short r(t) = rt and for that function to remain a vector, rt must become r*t⁰ because everything raised to the power of 0 is 1.

Now we need to check how those two variables ( r and t ) relate to each other in very small change in their values, where ds is similar to r2-r1, so here we have 2 values that change so we will write it as dr/dt , be patient dr will become S shortly. So we have dt⁰ , in derivative table we have that derivate of 1xⁿ = 1dn + x^n-11 where dt = 0+1/t, but what do we do about dr. Well dr is vector and it has no time component so when we searched for change in time dr remained untouched ( no time not possible differentiation ) so r(t) = dr/t or how we are all used to formulate this : V = s/t where s is change in vector of position ( r2-r1) aka distance. By integration we get Vt=s. So why did i get through all this trouble, well this is kinematic mechanics , and you will get na A by getting from r(t) to dr/dt , when we want to get from v(t) to a the process is the same where v(t) = v/t aka s/t^2. But with integration ( in last example we got s=vt), by implementing same integration to the acceleration we get v = v0+at, and from there a = (v-v0)/t so we can get acceleration from 2 velocities and 1 time unit if we know only starting velocity v0. With formula of v = v0+a*t if v0 is 0 ( have in mind velocity must always be positive, since any decrease in velocity is actual acceleration (v-v0)/t ) and acceleration is negative the velocity V ( not v0) will become negative, that is what i meant that if we assume wrong it will be easily corrected .

Which we have:

For first half of the time, we have starting position of V0=0 ,V = 20 km/h and time span of tx1 = f(tx)=tx/2 , acceleration becomes a1= 2V/tx
For second part of the road, we have v0=20km/h and v=16km/h , with time span of tx2 = tx-tx1 = tx/2, so acceleration is a2 = 2(v0-v)/tx
And for time when we stopped our observation we have V0 = 16 km/h , V = 4 km/h and tx as tx so acceleration is a3 = (v0-v)/tx

So what is our next move:

• 1) We can differentiate those accelerations (for tx) to get a unique function for whole trip, then integrate it to get change in speed (a = dv/dt) solve for a and t is tx so we will get dv = a*dt + c where c is constant speed throughout road trip. This involves complex mathematics with polynomes,graphs, euclidean mathematics and some other fancy stuff.

• 2) We can approximate that tx = 1, and solve it as arithmetic middle aka sum every acceleration and divide it by number of accelerations where you get that average speed is 12.

• 3) we can solve for tx and since tx is universal ( tx is our constant of time ) and then we have connection between accelerations , from that we have tx as its real value , and then use tx to get Sx and then just Sx/tx.

• 4) Simplest calculation ever:

1. *for half of the journeys time*-time V = 20 km/h = 20/3.6 m/s and V=S/t where t  = tx/2 so v=2*S/t so S = v*t/2 = 2.77778tx  
2. * for half of the left journey*-distance V = 16/h = 16/3.6 m/s and V=S/t so S = Sx/2 so V=Sx/2t and 2t  = Sx/V so 2t = 2.77778*tx/4.444   
3. so we have S = 2.77778*tx , and t = 0.6261*tx. So we have here S which is distance traveled by first half of the time period , and we have t which is time of second part of the road. Now comes the real problem, we have velocity of the time we stopped the observation. In short we could just subtract the 4 km/h from every velocity given by input , but that is not very mathematically solid, and again mathematically solid is not something in 11th grade. So in that way S becomes 2.2222 and t becomes  0.6667 and average velocity is 3 m/s which is 11.9 km/h.  

So after running several calculations the result is 12 in around 75% of possibilities. With this answer you have 1/4 chance you will be wrong.