Calculus (differentiation)?

. |
. |
A|
/|\
. |
. |80t
. |
P|____________B _ _ _ _
. .<--100-60t ------>
. .<------------100 ----------->

At any time t (0≦ t ≦ 1h),
distance of car A from P = 80t
distance of car A from P =100-60t
distance between A & B, m(AB)
= √ [(80t)² + (100-60t)²] . . . . . . . (from △APB, by Pythagoras Thm)
= 10√ [ 64t² + (10-6t)²]
= 10√ (100t² -120t +100)
= 20√ (25t² -30t +25)

∴ rate of change of distance s between A,B respect to time t, ds/dt
=d [20√ (25t² -30t +25)] dt
= 20*[(1/2) (25t² -30t +25)^ (-1/2)] (50t-30) . . . . . . . (by Chain Rule)
= 100*(5t-3)*(25t² -30t +25)^ (-1/2)

After 1 hr, ds/dt｜t=1
= 100*(2)*(25 -30 +25)^ (-1/2)
= 200 / √20
= 100 / √5
= 20√5 km/h