The answer is: #v_m=4m/s#
The average velocity is, by definition:
#v_m=(Deltas)/(Deltat)#,
where #Deltas# is the walked space and #Deltat# is the time taken.
We know that:
#v^2=v_0^2+2aDeltas#
where, in the first case:
#v# is the speed in the position #x=2m#
#v_0=# is the initial speed (in #x=0m#)
#a# is the given acceleration and #Deltas# is #2m-0m=2m#.
#v=sqrt(v_0^2+2aDeltas)=sqrt(0+2*4m/s^2*2m)=4m/s#
and, now, again, in the second case:
#v# is the speed in the position #x=8m#
#v_0=# is the initial speed (in #x=2m#)
#a# is the given acceleration and #Deltas# is #8m-2m=6m#.
#v=sqrt(v_0^2+2aDeltas)=sqrt(4^2m^2/s^2+2*4m/s^2*6m)=8m/s#.
Now let's find the time that is needed to go from #x=2# to #x=8#, using the rule:
#v=v_0+at#
in which:
#v# is the velocity in the position #x=8#
#v_0# is the velocity in the position #x=2#
#a# is the given acceleration.
#t=(v-v_0)/a=(8m/s-2m/s)/(4m/s^2)=6/4s=1.5s#.
So, finally:
#v_m=(8m-2m)/(1.5s)=6/1.5m/s=4m/s#.
