Basically, you want to take the integral of each interval and compare them. The two intervals with the same integral represent equal displacement of the particle. And since delta(x) is always 2, all you have to do is average the initial and final velocities of each interval and multiply by two to find total displacement.
Hope it helped.
Edit to show calculations:
2 * [ (0 + 10)/2 ] = 10 for interval AB
2 * [ (7 + 3)/2 ] = 10 for interval DE
Dew - small water drops, happens all the time, even to windows.
Let the distance between the towns be d and the speed of the air be s.
distance = speed * time
convert the minutes time into hours.
When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:
d
s−15
=
7
3
return trip is then :
d
s+15
=
7
5
Cross-multiplying both we get the two-variable system:
3d=7∗(s−15)5d=7∗(s+15)
3d=7s−1055d=7s+105
subtract first equation from second equation we get
2d=210d=105km
Substitute the value of d in the above equations for s.
5∗105=7s+1057s=420s=60km/hr
