Question

In a flight of 6000 km an aircraft was slowed down due to bad weather . Its average speed for the trip was reduced by
400km/hr and time increased by 30 min. Find the original duration of flight

Answer 1

Let Time = Distance / speed. Let time be 't ' and speed be 's ', then, t = 6000/s Distance = speed * time , 6000 = (s - 400)(t+1/2) (30 min. = 1/2 hour)=> 6000 = (s-400)(6000/s +1/2) (t = 6000/s)=> 12000s = (s-400)2s(6000/s + 1/2) (multiplying both sides 2s)=> 12000s=(s-400)(12000 + s) => 12000s = 12000s + s2 -4800000 - 400s=> 0 = s2 -400s - 4800000=> 0 =s2 - 2400s + 2000s - 4800000=> 0 = s (s- 2400 ) + 2000(s - 2400)=> 0 = (s + 2000)(s-2400)=> s= -2000, 2400Since, speed cannot be negative, we take it 2400km/h

WE have already know t = 6000/s

t = 6000/ 2400 = 5/2 = 2 + 1/2 ; t = 2 hours and 30 minutes