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Step-by-step explanation:
Let the usual speed of the plane is=s km/ hr
and the usual time it takes to reach the destination=t hr
Distance= 1500 km
Now, speed=distance/time
s=1500/t
t=1500/s-(I)
The plane ✈ got delayed by half an hour and speed was increased by 250 km/ hr to reach destination on time.
now, speed of plane=s+250 km/ hr
t = t -1/2hr
Now, speed × time = distance
(s+250) (t-1/2) =1500
(s+250) (1500/s-1/2) =1500 { from(I) we know t= 1500/t}
{s+250}{(3000-s)/2s}=1500
3000s-s^2+250×3000-250s = 3000s
-s^2+250×3000-250s=0
s^2+250s-250×3000=0
s^2+1000s-250s-250×3000=0
(s+1000) (s-750) =O
therefore,
s = -1000, s = +750
speed can't be negative.
so, usual speed = 750 km
or usual time = 1500/750
= 2 hr